Mixing
Relationship with derivative
$begingroup$
Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.
- $F(x)-F(y) = (x-y)F’(x)$
- $F(x)-F(y) geq (x-y)F’(x)$
- $F(x)-F(y) leq (x-y)F’(x)$
4.$F(x) -F(y) = F’(x) -F’(y)$
If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.
calculus limits
asked Jan 21 at 7:41
user601297user601297
36919
$endgroup$
add a comment |
$begingroup$
Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.
- $F(x)-F(y) = (x-y)F’(x)$
- $F(x)-F(y) geq (x-y)F’(x)$
- $F(x)-F(y) leq (x-y)F’(x)$
4.$F(x) -F(y) = F’(x) -F’(y)$
If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.
calculus limits
asked Jan 21 at 7:41
user601297user601297
36919
$endgroup$
add a comment |
$begingroup$
Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.
- $F(x)-F(y) = (x-y)F’(x)$
- $F(x)-F(y) geq (x-y)F’(x)$
- $F(x)-F(y) leq (x-y)F’(x)$
4.$F(x) -F(y) = F’(x) -F’(y)$
If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.
calculus limits
asked Jan 21 at 7:41
user601297user601297
36919
$endgroup$
Let $F :[0,1]rightarrow mathbb{R}$ be a differentiable function such that it’s derivative $F’(x)$ is increasing in $x$. Then which of the following is true for every $x$,$y$ in $[0,1]$ with $x>y$.
- $F(x)-F(y) = (x-y)F’(x)$
- $F(x)-F(y) geq (x-y)F’(x)$
- $F(x)-F(y) leq (x-y)F’(x)$
4.$F(x) -F(y) = F’(x) -F’(y)$
If I take the function to be $F(x) = x^2$ and take $x = 1$ and $y= 0$ then $(x-y)F’(x) = (1-0)F’(1) = 2$ which is greater than $F(1)-F(0) = 1$ , so I’m thinking that the correct option should be 3, but I am not sure about my approach, I feel like a formal proof would require using limits, but I couldn’t figure out how exactly to go about it, can someone provide a way to formally prove it please.
calculus limits
calculus limits
asked Jan 21 at 7:41
user601297user601297
36919
asked Jan 21 at 7:41
user601297user601297
36919
asked Jan 21 at 7:41
user601297user601297
36919
asked Jan 21 at 7:41
user601297user601297
36919
asked Jan 21 at 7:41
user601297user601297
36919
36919
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Credit to Kavi Rama Murthy.
MVT:
For $x>y:$
$F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.
Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$
Hence 1)-3)?
4) $f(t)=t^2$;
$f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.
$=f'(x)-f'(y)=2(x-y)$.
Now choose $x,y in (0,1)$(Why?) to rule out 4).
Appended:
Proof of option 3:
We have MVT:
$F(x)-F(y)= F'(a)(x-y) le$
$F'(x)(x-y)$,
since $F'(a)le F'(x)$, recall $y <a<x$.
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
add a comment |
$begingroup$
3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Credit to Kavi Rama Murthy.
MVT:
For $x>y:$
$F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.
Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$
Hence 1)-3)?
4) $f(t)=t^2$;
$f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.
$=f'(x)-f'(y)=2(x-y)$.
Now choose $x,y in (0,1)$(Why?) to rule out 4).
Appended:
Proof of option 3:
We have MVT:
$F(x)-F(y)= F'(a)(x-y) le$
$F'(x)(x-y)$,
since $F'(a)le F'(x)$, recall $y <a<x$.
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
add a comment |
$begingroup$
Credit to Kavi Rama Murthy.
MVT:
For $x>y:$
$F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.
Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$
Hence 1)-3)?
4) $f(t)=t^2$;
$f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.
$=f'(x)-f'(y)=2(x-y)$.
Now choose $x,y in (0,1)$(Why?) to rule out 4).
Appended:
Proof of option 3:
We have MVT:
$F(x)-F(y)= F'(a)(x-y) le$
$F'(x)(x-y)$,
since $F'(a)le F'(x)$, recall $y <a<x$.
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
add a comment |
$begingroup$
Credit to Kavi Rama Murthy.
MVT:
For $x>y:$
$F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.
Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$
Hence 1)-3)?
4) $f(t)=t^2$;
$f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.
$=f'(x)-f'(y)=2(x-y)$.
Now choose $x,y in (0,1)$(Why?) to rule out 4).
Appended:
Proof of option 3:
We have MVT:
$F(x)-F(y)= F'(a)(x-y) le$
$F'(x)(x-y)$,
since $F'(a)le F'(x)$, recall $y <a<x$.
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
Credit to Kavi Rama Murthy.
MVT:
For $x>y:$
$F(x)-F(y)= (x-y)F'(a)$ , $a in (y,x)$.
Since $F' $ is increasing: $F'(y)le F'(a) le F'(x).$
Hence 1)-3)?
4) $f(t)=t^2$;
$f(x)-f(y)=x^2-y^2= (x-y)(x+y)$.
$=f'(x)-f'(y)=2(x-y)$.
Now choose $x,y in (0,1)$(Why?) to rule out 4).
Appended:
Proof of option 3:
We have MVT:
$F(x)-F(y)= F'(a)(x-y) le$
$F'(x)(x-y)$,
since $F'(a)le F'(x)$, recall $y <a<x$.
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
edited Jan 21 at 15:49
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
answered Jan 21 at 8:48
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
add a comment |
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
Thanks for the answer, I was actually looking for a formal proof, like if I wasn’t given the options and the question was to proove 3. then how would you go about it.
$endgroup$
– user601297
Jan 21 at 9:54
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
user601297.Does this appended part answer your question?
$endgroup$
– Peter Szilas
Jan 21 at 10:30
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
The first line is throwing me off, $F(x)-F(y) =$ what?? There is nothing after the equal sign
$endgroup$
– user601297
Jan 21 at 11:50
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
$begingroup$
Ok after thinking about it for a while, I got the proof, thanks
$endgroup$
– user601297
Jan 21 at 12:45
add a comment |
$begingroup$
3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
add a comment |
$begingroup$
3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
add a comment |
$begingroup$
3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
$endgroup$
3) is true (by MVT and the fact that $f'$ is increasing) and all others are false. Consider the function $f(t)=t^{2}$ to see that other options are all false.
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
edited Jan 21 at 7:46
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
answered Jan 21 at 7:43
Kavi Rama MurthyKavi Rama Murthy
65.1k42766
65.1k42766
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
add a comment |
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
$begingroup$
Can you tell me the process to come to that conclusion?
$endgroup$
– user601297
Jan 21 at 7:44
add a comment |
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