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How to calculate $f^{primeprime}(3)$ from a value table of $f(x)$?
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Is there enough information to calculate $f^{primeprime}(3)$ from this table?
My intuition says there is a way.
begin{align}
f(3) &= 6\
f^prime(3) &= 1\
&= lim_{xto3}dfrac{f(x)-f(3)}{x-3}\
f^{primeprime} &= lim_{xto3}dfrac{f^prime(x)-f^prime(3)}{x-3}\
&= lim_{xto3}dfrac{f^prime(x)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-f(3)}{x-3}Bigg)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-6}{x-3}Bigg)-1}{x-3}\
&= Bigg(lim_{xto3}dfrac{f(x)-6}{(x-3)^2}Bigg)-lim_{xto3}dfrac{1}{x-3}\
end{align}
and then I get stuck...
calculus derivatives
asked 13 hours ago
kaisa
425
add a comment |
up vote
0
down vote
favorite
Is there enough information to calculate $f^{primeprime}(3)$ from this table?
My intuition says there is a way.
begin{align}
f(3) &= 6\
f^prime(3) &= 1\
&= lim_{xto3}dfrac{f(x)-f(3)}{x-3}\
f^{primeprime} &= lim_{xto3}dfrac{f^prime(x)-f^prime(3)}{x-3}\
&= lim_{xto3}dfrac{f^prime(x)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-f(3)}{x-3}Bigg)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-6}{x-3}Bigg)-1}{x-3}\
&= Bigg(lim_{xto3}dfrac{f(x)-6}{(x-3)^2}Bigg)-lim_{xto3}dfrac{1}{x-3}\
end{align}
and then I get stuck...
calculus derivatives
asked 13 hours ago
kaisa
425
In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there enough information to calculate $f^{primeprime}(3)$ from this table?
My intuition says there is a way.
begin{align}
f(3) &= 6\
f^prime(3) &= 1\
&= lim_{xto3}dfrac{f(x)-f(3)}{x-3}\
f^{primeprime} &= lim_{xto3}dfrac{f^prime(x)-f^prime(3)}{x-3}\
&= lim_{xto3}dfrac{f^prime(x)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-f(3)}{x-3}Bigg)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-6}{x-3}Bigg)-1}{x-3}\
&= Bigg(lim_{xto3}dfrac{f(x)-6}{(x-3)^2}Bigg)-lim_{xto3}dfrac{1}{x-3}\
end{align}
and then I get stuck...
calculus derivatives
asked 13 hours ago
kaisa
425
Is there enough information to calculate $f^{primeprime}(3)$ from this table?
My intuition says there is a way.
begin{align}
f(3) &= 6\
f^prime(3) &= 1\
&= lim_{xto3}dfrac{f(x)-f(3)}{x-3}\
f^{primeprime} &= lim_{xto3}dfrac{f^prime(x)-f^prime(3)}{x-3}\
&= lim_{xto3}dfrac{f^prime(x)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-f(3)}{x-3}Bigg)-1}{x-3}\
&= lim_{xto3}dfrac{Bigg(lim_{xto3}dfrac{f(x)-6}{x-3}Bigg)-1}{x-3}\
&= Bigg(lim_{xto3}dfrac{f(x)-6}{(x-3)^2}Bigg)-lim_{xto3}dfrac{1}{x-3}\
end{align}
and then I get stuck...
calculus derivatives
calculus derivatives
asked 13 hours ago
kaisa
425
asked 13 hours ago
kaisa
425
asked 13 hours ago
kaisa
425
asked 13 hours ago
kaisa
425
asked 13 hours ago
kaisa
425
425
In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago
add a comment |
In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago
In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago
add a comment |
3 Answers
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up vote
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With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:
Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.
For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.
For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).
What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get
$$ f''(3)simeq frac{1}{2}(-3-1) = -2$$
This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.
Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=lim_{xto 0}frac{g(x)-g(0)}{x-0}=lim_{xto 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).
It also follows that $f(x)$ is differentiable on $mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.
answered 13 hours ago
Taladris
4,62431832
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2
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If your $Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.
I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.
$f'(x)=limlimits_{c->0} frac{f(x+c)-f(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(x+c)-f'(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(4)-f'(2)}{2}=-2$
$f''(x)=limlimits_{c->0}frac{frac{f(x+2c)-f(x)}{2c}-frac{f(x)-f(x-2c)}{2c}}{2c}$
$f''(x)=limlimits_{c->0} frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$
All the $Delta x$ are =1 here. Let c=1.
$f''(3)=limlimits_{c->0} frac{f(5)-2f(3)+f(1)}{4(1)}=frac{-7}{4}$
So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.
edited 9 hours ago
Kemono Chen
1,550330
answered 13 hours ago
TurlocTheRed
57819
add a comment |
up vote
1
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This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
x + 3 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
frac16x^2 + frac92 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
for which $f''(3) = frac13$.
The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $lim_{h to 0}frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $frac{-3 + -1}{2} = -2$.
answered 13 hours ago
Reese
14.9k11136
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3 Answers
3
active
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:
Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.
For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.
For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).
What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get
$$ f''(3)simeq frac{1}{2}(-3-1) = -2$$
This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.
Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=lim_{xto 0}frac{g(x)-g(0)}{x-0}=lim_{xto 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).
It also follows that $f(x)$ is differentiable on $mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.
answered 13 hours ago
Taladris
4,62431832
add a comment |
up vote
1
down vote
accepted
With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:
Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.
For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.
For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).
What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get
$$ f''(3)simeq frac{1}{2}(-3-1) = -2$$
This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.
Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=lim_{xto 0}frac{g(x)-g(0)}{x-0}=lim_{xto 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).
It also follows that $f(x)$ is differentiable on $mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.
answered 13 hours ago
Taladris
4,62431832
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:
Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.
For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.
For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).
What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get
$$ f''(3)simeq frac{1}{2}(-3-1) = -2$$
This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.
Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=lim_{xto 0}frac{g(x)-g(0)}{x-0}=lim_{xto 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).
It also follows that $f(x)$ is differentiable on $mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.
answered 13 hours ago
Taladris
4,62431832
With the given information, it is not possible to determine $f''(3)$ exactly. Since only the values of $f$ near $a=3$ are relevant, I will ignore the other information and only consider $f(3)=6$ and $f'(3)=1$. Here are a few examples:
Consider $f(x)=x+3$. Then $f(3)=6$, $f'(3)=1$ and $f''(3)=0$.
For the quadratic function $f(x)=a(x-3)^2+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(3)=2a$ can be anything.
For $f(x)=(x-3)|x-3|+(x-3)+6$, we have $f(3)=6$, $f'(3)=1$ but $f''(x)$ does not exist (see below).
What can be done in your case is a broad estimation of $f''(3)$ assuming $f$ is "nice enough": the average rate of change of $f'$ between $2$ and $3$ is $f'_{[2,3]}=frac{f'(3)-f'(2)}{3-2}= -3$ and the average rate of change of $f'$ between $3$ and $4$ is $f'_{[3,4]}=frac{f'(4)-f'(3)}{4-3}= -1$. Average these two values, we get
$$ f''(3)simeq frac{1}{2}(-3-1) = -2$$
This comes from the definition of tangent line and derivatives as a limit of secant lines, but that's assuming quite a lot on the function $f$. And since $2$ and $4$ are not really "near $3$", that's probably a really bad estimation.
Study of $f(x)=(x-3)|x-3|+(x-3)+6$: Let $g(x)=x|x|$, so that $f(x)=g(x-3)+(x-3)+6$. For $x>0$, $g(x)=x^2$, so $g'(x)=2x$. Similarly, $g'(x)=-2x$ for $x<0$. For $x=0$, $g'(0)=lim_{xto 0}frac{g(x)-g(0)}{x-0}=lim_{xto 0}|x|=0$. To sum up, $g'(x)=2|x|$. It follows that $g''(0)$ does not exist (since $|x|$ has a cusp at $0$).
It also follows that $f(x)$ is differentiable on $mathbb R$ and $f'(x)=2|x-3|+1$, so $f'(3)=1$ (Chain Rule) abd $f''(x)$ is not differentiable at $3$.
answered 13 hours ago
Taladris
4,62431832
answered 13 hours ago
Taladris
4,62431832
answered 13 hours ago
Taladris
4,62431832
answered 13 hours ago
Taladris
4,62431832
4,62431832
add a comment |
add a comment |
up vote
2
down vote
If your $Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.
I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.
$f'(x)=limlimits_{c->0} frac{f(x+c)-f(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(x+c)-f'(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(4)-f'(2)}{2}=-2$
$f''(x)=limlimits_{c->0}frac{frac{f(x+2c)-f(x)}{2c}-frac{f(x)-f(x-2c)}{2c}}{2c}$
$f''(x)=limlimits_{c->0} frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$
All the $Delta x$ are =1 here. Let c=1.
$f''(3)=limlimits_{c->0} frac{f(5)-2f(3)+f(1)}{4(1)}=frac{-7}{4}$
So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.
edited 9 hours ago
Kemono Chen
1,550330
answered 13 hours ago
TurlocTheRed
57819
add a comment |
up vote
2
down vote
If your $Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.
I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.
$f'(x)=limlimits_{c->0} frac{f(x+c)-f(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(x+c)-f'(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(4)-f'(2)}{2}=-2$
$f''(x)=limlimits_{c->0}frac{frac{f(x+2c)-f(x)}{2c}-frac{f(x)-f(x-2c)}{2c}}{2c}$
$f''(x)=limlimits_{c->0} frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$
All the $Delta x$ are =1 here. Let c=1.
$f''(3)=limlimits_{c->0} frac{f(5)-2f(3)+f(1)}{4(1)}=frac{-7}{4}$
So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.
edited 9 hours ago
Kemono Chen
1,550330
answered 13 hours ago
TurlocTheRed
57819
add a comment |
up vote
2
down vote
up vote
2
down vote
If your $Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.
I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.
$f'(x)=limlimits_{c->0} frac{f(x+c)-f(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(x+c)-f'(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(4)-f'(2)}{2}=-2$
$f''(x)=limlimits_{c->0}frac{frac{f(x+2c)-f(x)}{2c}-frac{f(x)-f(x-2c)}{2c}}{2c}$
$f''(x)=limlimits_{c->0} frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$
All the $Delta x$ are =1 here. Let c=1.
$f''(3)=limlimits_{c->0} frac{f(5)-2f(3)+f(1)}{4(1)}=frac{-7}{4}$
So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.
edited 9 hours ago
Kemono Chen
1,550330
answered 13 hours ago
TurlocTheRed
57819
If your $Delta x$ can't get arbitrarily small you won't get a very accurate derivative. Many functions allow you to get close.
I think you can get close using Fibonacci Numbers. There are at least a couple possible approximations based on the table.
$f'(x)=limlimits_{c->0} frac{f(x+c)-f(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(x+c)-f'(x-c)}{2c}$
$f''(x)=limlimits_{c->0} frac{f'(4)-f'(2)}{2}=-2$
$f''(x)=limlimits_{c->0}frac{frac{f(x+2c)-f(x)}{2c}-frac{f(x)-f(x-2c)}{2c}}{2c}$
$f''(x)=limlimits_{c->0} frac{f(x+2c)-2f(x)+f(x-2c)}{4c^2}$
All the $Delta x$ are =1 here. Let c=1.
$f''(3)=limlimits_{c->0} frac{f(5)-2f(3)+f(1)}{4(1)}=frac{-7}{4}$
So these techniques give you approxmations of -2 and -7/4. Their average would probably be closer to the right value.
edited 9 hours ago
Kemono Chen
1,550330
answered 13 hours ago
TurlocTheRed
57819
edited 9 hours ago
Kemono Chen
1,550330
edited 9 hours ago
Kemono Chen
1,550330
edited 9 hours ago
Kemono Chen
1,550330
1,550330
answered 13 hours ago
TurlocTheRed
57819
answered 13 hours ago
TurlocTheRed
57819
answered 13 hours ago
TurlocTheRed
57819
57819
add a comment |
add a comment |
up vote
1
down vote
This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
x + 3 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
frac16x^2 + frac92 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
for which $f''(3) = frac13$.
The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $lim_{h to 0}frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $frac{-3 + -1}{2} = -2$.
answered 13 hours ago
Reese
14.9k11136
add a comment |
up vote
1
down vote
This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
x + 3 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
frac16x^2 + frac92 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
for which $f''(3) = frac13$.
The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $lim_{h to 0}frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $frac{-3 + -1}{2} = -2$.
answered 13 hours ago
Reese
14.9k11136
add a comment |
up vote
1
down vote
up vote
1
down vote
This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
x + 3 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
frac16x^2 + frac92 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
for which $f''(3) = frac13$.
The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $lim_{h to 0}frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $frac{-3 + -1}{2} = -2$.
answered 13 hours ago
Reese
14.9k11136
This isn't nearly enough information to be sure of the value of $f''(3)$. To take a simple example, $f$ could be the function
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
x + 3 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
This has $f''(x) = 0$ for all $x$. On the other hand, $f$ could also be
$$f(x) = begin{cases}
3x & text{for } x < 1.5 \
4x - 4 & text{for } 1.5 leq x < 2.5 \
frac16x^2 + frac92 & text{for } 2.5 leq x < 3.5 \
-1 & text{for } 3.5 leq x < 4.5 \
5x - 23 & text{for } 4.5 leq x
end{cases}$$
for which $f''(3) = frac13$.
The best you can possibly hope for here is an estimation to $f''(3)$. Recall that $f''(3)$ is $lim_{h to 0}frac{f'(3 + h) - f'(3)}{h}$. We only know a few values of $f'$, so we have to use those - it makes sense to use the smallest, $1$ and $-1$. For $h = 1$, $frac{f'(3 + h) - f'(3)}{h} = -1$. For $h = -1$, $frac{f'(3 + h) - f'(3)}{h} = -3$. So $-1$ and $-3$ would both be reasonable estimates. A somewhat more reasonable estimate would be the average between them: $frac{-3 + -1}{2} = -2$.
answered 13 hours ago
Reese
14.9k11136
answered 13 hours ago
Reese
14.9k11136
answered 13 hours ago
Reese
14.9k11136
answered 13 hours ago
Reese
14.9k11136
14.9k11136
add a comment |
add a comment |
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In general, it is not possible. It depends on the values of $f'(x)$ near $3$. It is even possible that $f'(x)$ is not defined near $3$ or $f''(3)$ is not defined. Do you have any additional information on $f$?
– Taladris
13 hours ago
@Taladris This is the only information I am given. Is this not enough information?
– kaisa
13 hours ago