Mixing
Prove or find a counterexample $forall x>0: f(2x)-f(x)<g(3x)-g(2x)$, given information about $f, g$.
$begingroup$
Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.
Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$
I've tried this for a long time, but I didn't make much progress.
I tried to move everything to one side and then take derivative.
If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.
I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).
Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.
Thanks.
real-analysis calculus analysis functions inequality
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
$endgroup$
add a comment |
$begingroup$
Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.
Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$
I've tried this for a long time, but I didn't make much progress.
I tried to move everything to one side and then take derivative.
If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.
I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).
Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.
Thanks.
real-analysis calculus analysis functions inequality
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
$endgroup$
2
$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55
add a comment |
$begingroup$
Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.
Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$
I've tried this for a long time, but I didn't make much progress.
I tried to move everything to one side and then take derivative.
If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.
I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).
Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.
Thanks.
real-analysis calculus analysis functions inequality
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
$endgroup$
Let $f,g:(0, infty) rightarrow mathbb{R}$ be two functions that satisfy the following for all $x>0$: $g'(x)>f'(x)$ and $f''(x)>0$.
Prove or find a counterexample:
$$
forall x>0: f(2x)-f(x)<g(3x)-g(2x).
$$
I've tried this for a long time, but I didn't make much progress.
I tried to move everything to one side and then take derivative.
If we move everything to the LHS and take derivative, we get $2f'(2x)-f'(x)+2g'(2x)-3g'(3x)$.
I don't know if this is positive or negative, and even if I knew, I don't see how it helps me. (It would help to know it if $f,g$ were defined at $0$).
Also I did not find any counterexample. Can someone please help me? And please write in your answer what was your intuition for this problem and why I still have no intuition to determine if the claim is true or false.
Thanks.
real-analysis calculus analysis functions inequality
real-analysis calculus analysis functions inequality
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
edited Jan 21 at 21:17
Daniele Tampieri
2,3972922
2,3972922
asked Jan 21 at 20:41
OmerOmer
3619
asked Jan 21 at 20:41
OmerOmer
3619
asked Jan 21 at 20:41
OmerOmer
3619
3619
2
$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55
add a comment |
2
$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55
2
2
$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55
$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$
There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$
So, we need to show that $f'(c)<g'(d).$
Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that
$$f'(c)<f'(d)<g'(d)$$ and we are done.
answered Jan 21 at 20:50
mflmfl
26.7k12142
$endgroup$
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$
There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$
So, we need to show that $f'(c)<g'(d).$
Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that
$$f'(c)<f'(d)<g'(d)$$ and we are done.
answered Jan 21 at 20:50
mflmfl
26.7k12142
$endgroup$
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
add a comment |
$begingroup$
There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$
There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$
So, we need to show that $f'(c)<g'(d).$
Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that
$$f'(c)<f'(d)<g'(d)$$ and we are done.
answered Jan 21 at 20:50
mflmfl
26.7k12142
$endgroup$
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
add a comment |
$begingroup$
There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$
There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$
So, we need to show that $f'(c)<g'(d).$
Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that
$$f'(c)<f'(d)<g'(d)$$ and we are done.
answered Jan 21 at 20:50
mflmfl
26.7k12142
$endgroup$
There exists $cin (x,2x)$ such that $$f(2x)-f(x)=f'(c)x.$$
There exists $din (2x,3x)$ such that $$g(3x)-g(2x)=g'(d)x.$$
So, we need to show that $f'(c)<g'(d).$
Now, $f''(x)>0$ implies that $f'(x)$ is increasing. So we have, using that $c<d$ and $f'(x)<g'(x),$ that
$$f'(c)<f'(d)<g'(d)$$ and we are done.
answered Jan 21 at 20:50
mflmfl
26.7k12142
answered Jan 21 at 20:50
mflmfl
26.7k12142
answered Jan 21 at 20:50
mflmfl
26.7k12142
answered Jan 21 at 20:50
mflmfl
26.7k12142
26.7k12142
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
add a comment |
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
$begingroup$
I like this answer. +$1$
$endgroup$
– Clayton
Jan 21 at 20:56
add a comment |
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$begingroup$
Intuition comes with experience. Give yourself two or three years and you'll develop the intuition.
$endgroup$
– Clayton
Jan 21 at 20:55