Prove that $f'(x) > frac {f(x)}{x}$ for a continuous, differentiable $f(x)$
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3
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I'm trying to prove the following:
Say $f:[0,1] to mathbb{R}$ is continuous on [0,1] and differentiable on (0,1). We also have $f(0)=0$ and $f'$ is strictly increasing. I want to prove that $f'(x) > frac {f(x)}{x}$.
So far I've tried proof by contradiction, so assume that $f'(x) leq frac {f(x)}{x}$. I'm guessing this assumption must either contradict $f$ being continuous, differentiable, or concave up, however, I can't seem to figure it out. Any pointers would be much appreciated!
calculus real-analysis
edited Nov 21 at 22:17
Frpzzd
20.1k638104
asked Nov 21 at 21:57
darcy
161
add a comment |
up vote
3
down vote
favorite
I'm trying to prove the following:
Say $f:[0,1] to mathbb{R}$ is continuous on [0,1] and differentiable on (0,1). We also have $f(0)=0$ and $f'$ is strictly increasing. I want to prove that $f'(x) > frac {f(x)}{x}$.
So far I've tried proof by contradiction, so assume that $f'(x) leq frac {f(x)}{x}$. I'm guessing this assumption must either contradict $f$ being continuous, differentiable, or concave up, however, I can't seem to figure it out. Any pointers would be much appreciated!
calculus real-analysis
edited Nov 21 at 22:17
Frpzzd
20.1k638104
asked Nov 21 at 21:57
darcy
161
The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm trying to prove the following:
Say $f:[0,1] to mathbb{R}$ is continuous on [0,1] and differentiable on (0,1). We also have $f(0)=0$ and $f'$ is strictly increasing. I want to prove that $f'(x) > frac {f(x)}{x}$.
So far I've tried proof by contradiction, so assume that $f'(x) leq frac {f(x)}{x}$. I'm guessing this assumption must either contradict $f$ being continuous, differentiable, or concave up, however, I can't seem to figure it out. Any pointers would be much appreciated!
calculus real-analysis
edited Nov 21 at 22:17
Frpzzd
20.1k638104
asked Nov 21 at 21:57
darcy
161
I'm trying to prove the following:
Say $f:[0,1] to mathbb{R}$ is continuous on [0,1] and differentiable on (0,1). We also have $f(0)=0$ and $f'$ is strictly increasing. I want to prove that $f'(x) > frac {f(x)}{x}$.
So far I've tried proof by contradiction, so assume that $f'(x) leq frac {f(x)}{x}$. I'm guessing this assumption must either contradict $f$ being continuous, differentiable, or concave up, however, I can't seem to figure it out. Any pointers would be much appreciated!
calculus real-analysis
calculus real-analysis
edited Nov 21 at 22:17
Frpzzd
20.1k638104
asked Nov 21 at 21:57
darcy
161
edited Nov 21 at 22:17
Frpzzd
20.1k638104
asked Nov 21 at 21:57
darcy
161
edited Nov 21 at 22:17
Frpzzd
20.1k638104
edited Nov 21 at 22:17
Frpzzd
20.1k638104
edited Nov 21 at 22:17
Frpzzd
20.1k638104
20.1k638104
asked Nov 21 at 21:57
darcy
161
asked Nov 21 at 21:57
darcy
161
asked Nov 21 at 21:57
darcy
161
161
The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18
add a comment |
The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18
The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18
add a comment |
2 Answers
2
active
oldest
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up vote
5
down vote
Due to mean value theorem exists $xiin ]0, x[$ such that
$$
f(x)=f(x)-0=f(x)-f(0)= xf'(xi)<xf'(x)
$$
because $xi<x$.
answered Nov 21 at 22:21
P De Donato
3147
add a comment |
up vote
1
down vote
Since $f'$ is strictly increasing, you have that $f'(a)lt f'(b)$ for all $alt b$, which implies that
$$f(x)=int_0^x f'(t)dtlt int_0^x f'(x)dtlt xf'(x)$$
so you have that $f(x)lt xf'(x)$ and you are done.
answered Nov 21 at 22:10
Frpzzd
20.1k638104
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Due to mean value theorem exists $xiin ]0, x[$ such that
$$
f(x)=f(x)-0=f(x)-f(0)= xf'(xi)<xf'(x)
$$
because $xi<x$.
answered Nov 21 at 22:21
P De Donato
3147
add a comment |
up vote
5
down vote
Due to mean value theorem exists $xiin ]0, x[$ such that
$$
f(x)=f(x)-0=f(x)-f(0)= xf'(xi)<xf'(x)
$$
because $xi<x$.
answered Nov 21 at 22:21
P De Donato
3147
add a comment |
up vote
5
down vote
up vote
5
down vote
Due to mean value theorem exists $xiin ]0, x[$ such that
$$
f(x)=f(x)-0=f(x)-f(0)= xf'(xi)<xf'(x)
$$
because $xi<x$.
answered Nov 21 at 22:21
P De Donato
3147
Due to mean value theorem exists $xiin ]0, x[$ such that
$$
f(x)=f(x)-0=f(x)-f(0)= xf'(xi)<xf'(x)
$$
because $xi<x$.
answered Nov 21 at 22:21
P De Donato
3147
answered Nov 21 at 22:21
P De Donato
3147
answered Nov 21 at 22:21
P De Donato
3147
answered Nov 21 at 22:21
P De Donato
3147
3147
add a comment |
add a comment |
up vote
1
down vote
Since $f'$ is strictly increasing, you have that $f'(a)lt f'(b)$ for all $alt b$, which implies that
$$f(x)=int_0^x f'(t)dtlt int_0^x f'(x)dtlt xf'(x)$$
so you have that $f(x)lt xf'(x)$ and you are done.
answered Nov 21 at 22:10
Frpzzd
20.1k638104
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
add a comment |
up vote
1
down vote
Since $f'$ is strictly increasing, you have that $f'(a)lt f'(b)$ for all $alt b$, which implies that
$$f(x)=int_0^x f'(t)dtlt int_0^x f'(x)dtlt xf'(x)$$
so you have that $f(x)lt xf'(x)$ and you are done.
answered Nov 21 at 22:10
Frpzzd
20.1k638104
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
add a comment |
up vote
1
down vote
up vote
1
down vote
Since $f'$ is strictly increasing, you have that $f'(a)lt f'(b)$ for all $alt b$, which implies that
$$f(x)=int_0^x f'(t)dtlt int_0^x f'(x)dtlt xf'(x)$$
so you have that $f(x)lt xf'(x)$ and you are done.
answered Nov 21 at 22:10
Frpzzd
20.1k638104
Since $f'$ is strictly increasing, you have that $f'(a)lt f'(b)$ for all $alt b$, which implies that
$$f(x)=int_0^x f'(t)dtlt int_0^x f'(x)dtlt xf'(x)$$
so you have that $f(x)lt xf'(x)$ and you are done.
answered Nov 21 at 22:10
Frpzzd
20.1k638104
edited Nov 21 at 22:20
answered Nov 21 at 22:10
Frpzzd
20.1k638104
answered Nov 21 at 22:10
Frpzzd
20.1k638104
answered Nov 21 at 22:10
Frpzzd
20.1k638104
20.1k638104
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
add a comment |
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
That's fine (if $f'$ is strictly increasing), but I'll bet that this exercise appears before the students have encountered integration, as it can be addressed simply with something like the mean value theorem.
– John Hughes
Nov 21 at 22:19
add a comment |
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The inequality you want to prove is a rearrangement of $[f(x)/x]'>0$. Maybe that helps?
– Arthur
Nov 21 at 22:06
The claim is false for $f(x) = x$. You want to have $f'$ strictly increasing.
– daw
Nov 21 at 22:12
@Arthur I did manage to get there by rearranging, but I'm not sure where to go from there. It seems like I need MVT but I'm still having trouble
– darcy
Nov 21 at 23:18