Let $f$ be differentiable for all $x$, $f(0)=0$, & $exists a,b>0$ such that $ale f'(x)le b$ all...
$begingroup$
Looking for hints on this, I have worked out that we know $f'(x)>0$ so $f$ is strictly increasing for all $xge0$ and I assume we need to use somewhere $f(0)=0$ but I do not know how to begin
My application of mean value theorem:
Apply mean value theorem for $f$ on the interval from $[0,x]$
There exists a $cin (0,x)$ such that $f'(c)=frac{f(x)-f(0)}{x-0}=frac{f(x)}{x}$
Hence,
$ale f'(x)le brightarrow ale frac{f(x)}{x}le b$ for $c,x>0$ and it follows $axle f(x)le bx$ for $c,x>0$
Lastly, in the case $x=c=0$ it is trivial.
■
real-analysis
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
$endgroup$
|
show 1 more comment
$begingroup$
Looking for hints on this, I have worked out that we know $f'(x)>0$ so $f$ is strictly increasing for all $xge0$ and I assume we need to use somewhere $f(0)=0$ but I do not know how to begin
My application of mean value theorem:
Apply mean value theorem for $f$ on the interval from $[0,x]$
There exists a $cin (0,x)$ such that $f'(c)=frac{f(x)-f(0)}{x-0}=frac{f(x)}{x}$
Hence,
$ale f'(x)le brightarrow ale frac{f(x)}{x}le b$ for $c,x>0$ and it follows $axle f(x)le bx$ for $c,x>0$
Lastly, in the case $x=c=0$ it is trivial.
■
real-analysis
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
$endgroup$
$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
1
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
2
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52
|
show 1 more comment
$begingroup$
Looking for hints on this, I have worked out that we know $f'(x)>0$ so $f$ is strictly increasing for all $xge0$ and I assume we need to use somewhere $f(0)=0$ but I do not know how to begin
My application of mean value theorem:
Apply mean value theorem for $f$ on the interval from $[0,x]$
There exists a $cin (0,x)$ such that $f'(c)=frac{f(x)-f(0)}{x-0}=frac{f(x)}{x}$
Hence,
$ale f'(x)le brightarrow ale frac{f(x)}{x}le b$ for $c,x>0$ and it follows $axle f(x)le bx$ for $c,x>0$
Lastly, in the case $x=c=0$ it is trivial.
■
real-analysis
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
$endgroup$
Looking for hints on this, I have worked out that we know $f'(x)>0$ so $f$ is strictly increasing for all $xge0$ and I assume we need to use somewhere $f(0)=0$ but I do not know how to begin
My application of mean value theorem:
Apply mean value theorem for $f$ on the interval from $[0,x]$
There exists a $cin (0,x)$ such that $f'(c)=frac{f(x)-f(0)}{x-0}=frac{f(x)}{x}$
Hence,
$ale f'(x)le brightarrow ale frac{f(x)}{x}le b$ for $c,x>0$ and it follows $axle f(x)le bx$ for $c,x>0$
Lastly, in the case $x=c=0$ it is trivial.
■
real-analysis
real-analysis
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
edited Dec 6 '18 at 18:29
Albert Diaz
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
asked Dec 6 '18 at 18:14
Albert DiazAlbert Diaz
1156
1156
$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
1
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
2
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52
|
show 1 more comment
$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
1
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
2
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52
$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
1
1
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
2
2
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52
|
show 1 more comment
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$begingroup$
Mean value theorem or fundamental theorem of calculus will do it
$endgroup$
– user25959
Dec 6 '18 at 18:15
$begingroup$
Suppose not. Then use the mean value theorem
$endgroup$
– davidlowryduda♦
Dec 6 '18 at 18:16
1
$begingroup$
I would use the monotonicity of integration.
$endgroup$
– Josh B.
Dec 6 '18 at 18:18
2
$begingroup$
@user25959 MVT yes. FTC no! The FTC has hypotheses that are not satisfied here...
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:50
$begingroup$
@JoshB. Nope. Given just that $f$ is differentiable it does not follow that $f$ is the integral of $f'$, so properties of integration are nnot going to help.
$endgroup$
– David C. Ullrich
Dec 6 '18 at 18:52