Mixing
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If $f(x)=int_{0}^{x}sqrt{f(t)},dt$ then $f(6)$ is?
$begingroup$
let $f:[0,infty) rightarrow[0,infty]$ be continuous on $[0,infty]$
and differentiable on $(0,infty)$. if $f(x)=int_{0}^{x}sqrt{f(t)}dt$
then $f(6)$ is ?
$f(0)=0$
$f'(x)=sqrt{f(x)}implies f'(6)=sqrt{f(6)}implies f'(6)^2=f(6)$
Now I am trying to find out $f'(6) $ somehow
$f'(6)=lim_{x rightarrow 6}dfrac{f(x)-f(6)}{x-6}=lim_{x rightarrow 6}dfrac{f(x)-f'(6)^2}{x-6}$
$f'(0)=lim_{x rightarrow 0}dfrac{f(x)-f(0)}{x-0}$
I am missing something and I am not able to solve this problem by above steps.
Any hint?
real-analysis calculus
asked Jan 26 at 6:20
Daman deepDaman deep
756419
$endgroup$
|
show 2 more comments
$begingroup$
let $f:[0,infty) rightarrow[0,infty]$ be continuous on $[0,infty]$
and differentiable on $(0,infty)$. if $f(x)=int_{0}^{x}sqrt{f(t)}dt$
then $f(6)$ is ?
$f(0)=0$
$f'(x)=sqrt{f(x)}implies f'(6)=sqrt{f(6)}implies f'(6)^2=f(6)$
Now I am trying to find out $f'(6) $ somehow
$f'(6)=lim_{x rightarrow 6}dfrac{f(x)-f(6)}{x-6}=lim_{x rightarrow 6}dfrac{f(x)-f'(6)^2}{x-6}$
$f'(0)=lim_{x rightarrow 0}dfrac{f(x)-f(0)}{x-0}$
I am missing something and I am not able to solve this problem by above steps.
Any hint?
real-analysis calculus
asked Jan 26 at 6:20
Daman deepDaman deep
756419
$endgroup$
1
$begingroup$
The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
$endgroup$
– alex.jordan
Jan 26 at 6:37
$begingroup$
@alex.jordan my bad
$endgroup$
– Daman deep
Jan 26 at 6:41
1
$begingroup$
Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
$endgroup$
– Clement C.
Jan 26 at 6:49
$begingroup$
As of right now, this question had 0 downvotes.
$endgroup$
– Chase Ryan Taylor
Jan 26 at 15:26
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@ChaseRyanTaylor What do you mean ?
$endgroup$
– Daman deep
Jan 26 at 15:29
|
show 2 more comments
$begingroup$
let $f:[0,infty) rightarrow[0,infty]$ be continuous on $[0,infty]$
and differentiable on $(0,infty)$. if $f(x)=int_{0}^{x}sqrt{f(t)}dt$
then $f(6)$ is ?
$f(0)=0$
$f'(x)=sqrt{f(x)}implies f'(6)=sqrt{f(6)}implies f'(6)^2=f(6)$
Now I am trying to find out $f'(6) $ somehow
$f'(6)=lim_{x rightarrow 6}dfrac{f(x)-f(6)}{x-6}=lim_{x rightarrow 6}dfrac{f(x)-f'(6)^2}{x-6}$
$f'(0)=lim_{x rightarrow 0}dfrac{f(x)-f(0)}{x-0}$
I am missing something and I am not able to solve this problem by above steps.
Any hint?
real-analysis calculus
asked Jan 26 at 6:20
Daman deepDaman deep
756419
$endgroup$
let $f:[0,infty) rightarrow[0,infty]$ be continuous on $[0,infty]$
and differentiable on $(0,infty)$. if $f(x)=int_{0}^{x}sqrt{f(t)}dt$
then $f(6)$ is ?
$f(0)=0$
$f'(x)=sqrt{f(x)}implies f'(6)=sqrt{f(6)}implies f'(6)^2=f(6)$
Now I am trying to find out $f'(6) $ somehow
$f'(6)=lim_{x rightarrow 6}dfrac{f(x)-f(6)}{x-6}=lim_{x rightarrow 6}dfrac{f(x)-f'(6)^2}{x-6}$
$f'(0)=lim_{x rightarrow 0}dfrac{f(x)-f(0)}{x-0}$
I am missing something and I am not able to solve this problem by above steps.
Any hint?
real-analysis calculus
real-analysis calculus
asked Jan 26 at 6:20
Daman deepDaman deep
756419
asked Jan 26 at 6:20
Daman deepDaman deep
756419
edited Jan 27 at 11:45
Daman deep
asked Jan 26 at 6:20
Daman deepDaman deep
756419
asked Jan 26 at 6:20
Daman deepDaman deep
756419
asked Jan 26 at 6:20
Daman deepDaman deep
756419
756419
1
$begingroup$
The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
$endgroup$
– alex.jordan
Jan 26 at 6:37
$begingroup$
@alex.jordan my bad
$endgroup$
– Daman deep
Jan 26 at 6:41
1
$begingroup$
Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
$endgroup$
– Clement C.
Jan 26 at 6:49
$begingroup$
As of right now, this question had 0 downvotes.
$endgroup$
– Chase Ryan Taylor
Jan 26 at 15:26
$begingroup$
@ChaseRyanTaylor What do you mean ?
$endgroup$
– Daman deep
Jan 26 at 15:29
|
show 2 more comments
1
$begingroup$
The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
$endgroup$
– alex.jordan
Jan 26 at 6:37
$begingroup$
@alex.jordan my bad
$endgroup$
– Daman deep
Jan 26 at 6:41
1
$begingroup$
Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
$endgroup$
– Clement C.
Jan 26 at 6:49
$begingroup$
As of right now, this question had 0 downvotes.
$endgroup$
– Chase Ryan Taylor
Jan 26 at 15:26
$begingroup$
@ChaseRyanTaylor What do you mean ?
$endgroup$
– Daman deep
Jan 26 at 15:29
1
1
$begingroup$
The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
$endgroup$
– alex.jordan
Jan 26 at 6:37
$begingroup$
The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
$endgroup$
– alex.jordan
Jan 26 at 6:37
$begingroup$
@alex.jordan my bad
$endgroup$
– Daman deep
Jan 26 at 6:41
$begingroup$
@alex.jordan my bad
$endgroup$
– Daman deep
Jan 26 at 6:41
1
1
$begingroup$
Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
$endgroup$
– Clement C.
Jan 26 at 6:49
$begingroup$
Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
$endgroup$
– Clement C.
Jan 26 at 6:49
$begingroup$
As of right now, this question had 0 downvotes.
$endgroup$
– Chase Ryan Taylor
Jan 26 at 15:26
$begingroup$
As of right now, this question had 0 downvotes.
$endgroup$
– Chase Ryan Taylor
Jan 26 at 15:26
$begingroup$
@ChaseRyanTaylor What do you mean ?
$endgroup$
– Daman deep
Jan 26 at 15:29
$begingroup$
@ChaseRyanTaylor What do you mean ?
$endgroup$
– Daman deep
Jan 26 at 15:29
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Assume $f(x) > 0$. Upon differentiation, we have $y' = sqrt{y} $ where $y=f(x)$. We have $int frac{dy }{y^{1/2}} = x + C $. Thus $2 sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
|
show 10 more comments
$begingroup$
Suppose $f(x) = 0$ for some $x in (0,infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = inf{y : f(y) = 0}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.
KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $sqrt f$.
Since $f' = sqrt f$, $f'$ is differentiable in $(x^*,infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (sqrt f)' = frac{1}{2sqrt f} times f' = frac 12$ everywhere on $(x^*,infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$.
Thus, $f''$ is constant on $(x^*,infty)$ and equal to $frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = frac 12 implies f = frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.
For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = mathbf 1_{x > x^*} frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $mathbf 1_{6 > x^*} frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$, which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $sqrt f$ to be invertible.
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
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$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
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– Jimmy Sabater
Jan 26 at 7:38
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True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
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– астон вілла олоф мэллбэрг
Jan 26 at 7:40
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I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
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– Jimmy Sabater
Jan 26 at 7:41
1
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Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
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– астон вілла олоф мэллбэрг
Jan 26 at 7:44
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Thank you for writing a correct answer to the question.
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– Clement C.
Jan 26 at 13:58
|
show 1 more comment
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An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=frac12$. Integrating once, we get $f’(x)=frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=frac{x^2}{4}.$$
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
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1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
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– Daman deep
Jan 26 at 13:57
1
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@Damandeep There is already one answer doing it.
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– Clement C.
Jan 26 at 13:58
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@Daman: I am in agreement.
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– Clayton
Jan 26 at 13:58
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
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– Clement C.
Jan 26 at 14:02
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Assume $f(x) > 0$. Upon differentiation, we have $y' = sqrt{y} $ where $y=f(x)$. We have $int frac{dy }{y^{1/2}} = x + C $. Thus $2 sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
|
show 10 more comments
$begingroup$
Assume $f(x) > 0$. Upon differentiation, we have $y' = sqrt{y} $ where $y=f(x)$. We have $int frac{dy }{y^{1/2}} = x + C $. Thus $2 sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
|
show 10 more comments
$begingroup$
Assume $f(x) > 0$. Upon differentiation, we have $y' = sqrt{y} $ where $y=f(x)$. We have $int frac{dy }{y^{1/2}} = x + C $. Thus $2 sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
$endgroup$
Assume $f(x) > 0$. Upon differentiation, we have $y' = sqrt{y} $ where $y=f(x)$. We have $int frac{dy }{y^{1/2}} = x + C $. Thus $2 sqrt{y} = x + C $ And $C=0$ as $f(0)=0$. Thus, $y = frac{x^2}{4}$. It follows that $f(6) = 9$
Added: See Clement C's comments below for a more rigorous solutions when $f(x) > 0$ is not assumed.
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
edited Jan 26 at 7:02
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
answered Jan 26 at 6:40
Jimmy SabaterJimmy Sabater
3,054325
3,054325
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
|
show 10 more comments
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
How do you argue that division by $sqrt{f(x)}$? You may want to explain why $f>0$, to avoid division by $0$.
$endgroup$
– Clement C.
Jan 26 at 6:43
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
$begingroup$
What if f(t) is identically zero.... U should deal that case too!
$endgroup$
– Cloud JR
Jan 26 at 6:46
3
3
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
$begingroup$
The diferential equation has not a uníque solution. For instance, $f(t)=0$ is also a solution. Al you can say is $0le f(6)le9$.
$endgroup$
– Julián Aguirre
Jan 26 at 6:46
1
1
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
$begingroup$
Actually, $f$ could be $$f(x) = begin{cases}0 & x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any $x_0 geq 0$.
$endgroup$
– Clement C.
Jan 26 at 6:47
1
1
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
$begingroup$
@clement C. Could you explain that
$endgroup$
– Cloud JR
Jan 26 at 6:48
|
show 10 more comments
$begingroup$
Suppose $f(x) = 0$ for some $x in (0,infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = inf{y : f(y) = 0}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.
KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $sqrt f$.
Since $f' = sqrt f$, $f'$ is differentiable in $(x^*,infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (sqrt f)' = frac{1}{2sqrt f} times f' = frac 12$ everywhere on $(x^*,infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$.
Thus, $f''$ is constant on $(x^*,infty)$ and equal to $frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = frac 12 implies f = frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.
For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = mathbf 1_{x > x^*} frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $mathbf 1_{6 > x^*} frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$, which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $sqrt f$ to be invertible.
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
1
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
|
show 1 more comment
$begingroup$
Suppose $f(x) = 0$ for some $x in (0,infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = inf{y : f(y) = 0}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.
KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $sqrt f$.
Since $f' = sqrt f$, $f'$ is differentiable in $(x^*,infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (sqrt f)' = frac{1}{2sqrt f} times f' = frac 12$ everywhere on $(x^*,infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$.
Thus, $f''$ is constant on $(x^*,infty)$ and equal to $frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = frac 12 implies f = frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.
For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = mathbf 1_{x > x^*} frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $mathbf 1_{6 > x^*} frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$, which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $sqrt f$ to be invertible.
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
1
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
|
show 1 more comment
$begingroup$
Suppose $f(x) = 0$ for some $x in (0,infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = inf{y : f(y) = 0}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.
KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $sqrt f$.
Since $f' = sqrt f$, $f'$ is differentiable in $(x^*,infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (sqrt f)' = frac{1}{2sqrt f} times f' = frac 12$ everywhere on $(x^*,infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$.
Thus, $f''$ is constant on $(x^*,infty)$ and equal to $frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = frac 12 implies f = frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.
For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = mathbf 1_{x > x^*} frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $mathbf 1_{6 > x^*} frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$, which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $sqrt f$ to be invertible.
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
$endgroup$
Suppose $f(x) = 0$ for some $x in (0,infty)$. Then, $f$ is the integral of a non-negative function from $0$ to $x$, so this forces $f(y) = 0$ for all $y leq x$.
On the contrary, $f$ being the integral of a non-negative function is increasing. Therefore, if $x^* = inf{y : f(y) = 0}$, then $f$ is positive beyond $x^*$ and zero before (and including) that point.
KEY POINT : Once $f$ is shown to be positive, so is the square root, and therefore the derivative exists since we may divide by $sqrt f$.
Since $f' = sqrt f$, $f'$ is differentiable in $(x^*,infty)$ (and positive on this interval, since we are considering the positive square root), with $f'' = (sqrt f)' = frac{1}{2sqrt f} times f' = frac 12$ everywhere on $(x^*,infty)$. Note that $f''(x^*) = 0$, from the fact that $f'$ is zero on $[0,x^*]$.
Thus, $f''$ is constant on $(x^*,infty)$ and equal to $frac 12$.We may solve this equation and obtain $f''$ as some quadratic on this interval, since $f '' = frac 12 implies f = frac{x^2}{4} + c_1x+c_2$, with the conditions at $x^*$. You can find what $c_1,c_2$ are.
Thus, if $f$ is as above, then we may characterize $f$ as being $0$ up till some $x^*$ and then the quadratic we obtained above. In particular, $f(6)$ does not take one svalue, but a range of values depending on which $f$ we are referring to. I am sure you will be able to deduce the exact range of values it lies in from here.
For completeness, I add the solution : For any $x^*$, the function $f_{x^*}(x) = mathbf 1_{x > x^*} frac{(x-x^*)^2}{4}$ satisfies the given conditions. Conversely, any function satisfying the given conditions must be of this form.
From here, one sees that $f(6)$ may take any of the values $mathbf 1_{6 > x^*} frac{(6-x^*)^2}{4}$, from where the decreasing nature of this function in $x^*$ gives the answer as $f(6) in [0,9]$. Note that if $x^* = 0$ then $f(6) = 9$, which is the "standard" answer under the common mistake of not realizing that $f$ needs to be positive for $sqrt f$ to be invertible.
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
edited Jan 26 at 7:27
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
answered Jan 26 at 7:13
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
39.8k33477
39.8k33477
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
1
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
|
show 1 more comment
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
1
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
This answer is quite sophisticated for the level of the OP. Just to point out something that the OP may have not seen, the notation ${bf 1_A}$. Also, given that the OP asked a question is single variable calculus, he definitely not seen infimum or supremums, etc, etc
$endgroup$
– Jimmy Sabater
Jan 26 at 7:38
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
True. If the OP would like, I can expand on these notions. However, for a complete answer we must delve into these notions.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:40
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
$begingroup$
I like the answer and I give +1 to this answer, but I doubt it will help the OP. But, anyway, it will help other people see a complete rigorous solution and that also matters.
$endgroup$
– Jimmy Sabater
Jan 26 at 7:41
1
1
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Given that your answer has been accepted (and I have up voted it as well!) I suppose your answer is the most useful to the OP, which makes good sense as well, since despite being incomplete it captures the main(intended) point.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 26 at 7:44
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
Thank you for writing a correct answer to the question.
$endgroup$
– Clement C.
Jan 26 at 13:58
|
show 1 more comment
$begingroup$
An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=frac12$. Integrating once, we get $f’(x)=frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=frac{x^2}{4}.$$
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
$endgroup$
1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
1
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
add a comment |
$begingroup$
An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=frac12$. Integrating once, we get $f’(x)=frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=frac{x^2}{4}.$$
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
$endgroup$
1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
1
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
add a comment |
$begingroup$
An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=frac12$. Integrating once, we get $f’(x)=frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=frac{x^2}{4}.$$
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
$endgroup$
An alternative method (the case $f=0$ is trivial; see @ClementC’s comment for slight variations of this answer where $f>0$ isn’t assumed for $x>0$):
You already know $f’(x)=sqrt{f(x)}$. Since the composition of differentials functions is differentiable (this is where we use the fact that $f(x)>0$ for $x>0$), we deduce that $f’’(x)=frac12$. Integrating once, we get $f’(x)=frac{x}{2}+C$, and using the fact that $f(0)=0$, we can see that $f’(0)=0$, meaning $C=0$. Integrating once more and using the initial value, we have $$f(x)=frac{x^2}{4}.$$
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
edited Jan 26 at 15:31
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
answered Jan 26 at 6:49
ClaytonClayton
19.5k33287
19.5k33287
1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
1
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
add a comment |
1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
1
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
1
1
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
$begingroup$
@ClementC. sir you should write an answer explaining all the things.
$endgroup$
– Daman deep
Jan 26 at 13:57
1
1
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Damandeep There is already one answer doing it.
$endgroup$
– Clement C.
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
$begingroup$
@Daman: I am in agreement.
$endgroup$
– Clayton
Jan 26 at 13:58
1
1
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
$begingroup$
@Clayton yes. in view of your latest edit, I'm removing my comments not to distract from your answer. (To be clear, I didn't care at all about credit! Just so that people seeing this question in the future can have a full correct answer (which, e.g., the currently accepted answer is not)
$endgroup$
– Clement C.
Jan 26 at 14:02
add a comment |
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The title doesn't match the question. Is the integrand $sqrt{t}$ or $sqrt{f(t)}$?
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– alex.jordan
Jan 26 at 6:37
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@alex.jordan my bad
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– Daman deep
Jan 26 at 6:41
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Without more details, $f$ could be any function of the form $$f(x) = begin{cases}0 & 0leq x leq x_0\ frac{(x-x_0)^2}{4}& x > x_0end{cases}$$ for any fixed $x_0 geq 0$. This means $f(6)$ can literally be any value between $0$ (for $x_0geq 6$) to $9$ (for $x_0 = 0$).
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– Clement C.
Jan 26 at 6:49
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As of right now, this question had 0 downvotes.
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– Chase Ryan Taylor
Jan 26 at 15:26
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@ChaseRyanTaylor What do you mean ?
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– Daman deep
Jan 26 at 15:29