# Thread: Show that g is differentiable and bounded

1. ## Show that g is differentiable and bounded

Given the piecewise function

g(x) =

{ x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
{ 0 if x = 0

Prove that g is differentiable and g' is bounded on (-infinity,infinity).

I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

This is the first time I have ever seen anything like this.

Thanks for any help.

2. Originally Posted by katielaw
Given the piecewise function

g(x) =

{ x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
{ 0 if x = 0

Prove that g is differentiable and g' is bounded on (-infinity,infinity).

I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

This is the first time I have ever seen anything like this.

Thanks for any help.
$g(x)=\left\{\begin{array}{lr}x^3e^{-x^2/4}\sin(4x^{-2}):&x\neq0\\0:&x=0\end{array}\right\}$

$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})$

Do you know where to go from here?

See Tonio's post below for proving $g'$ is bounded.

3. Originally Posted by katielaw
Given the piecewise function

g(x) =

{ x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
{ 0 if x = 0

Prove that g is differentiable and g' is bounded on (-infinity,infinity).

I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

This is the first time I have ever seen anything like this.

Thanks for any help.

$g(x)=\left\{\begin{array}{cc}x^3e^{-x^2\slash 4}\sin\frac{4}{x^2}&\,\mbox{ if }\,x\neq 0\\0&\,\mbox{ if }\,x=0\end{array}\right.$

$g'(x)=\left\{\begin{array}{cc}e^{-x^2\slash 4}\left(3x^2\sin\frac{4}{x^2}-\frac{x^4}{2}\sin\frac{4}{x^2}-8\cos \frac{4}{x^2}\right)&\,\mbox{ if }\,x\neq 0\\4&\,\mbox{ if }\,x=0\end{array}\right.$

Pay attention to the fact that the derivative is an even function and thus it's enough to show boundness in, say $(0,\infty)$,:

$|g'(x)|=\left|e^{-x^2\slash 4}\left(3x^2\sin\frac{4}{x^2}-\frac{x^4}{2}\sin\frac{4}{x^2}-8\cos \frac{4}{x^2}\right)\right|=$ $\left|e^{-x^2\slash 4}\right|\left|12\,\frac{\sin\frac{4}{x^2}}{\frac{ 4}{x^2}}-2x^2\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}-8\cos\frac{4}{x^2}\right|$

But $12\,\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}\xright arrow [x\to \infty] {}12$ , $-8\cos\frac{4}{x^2}\xrightarrow [x\to \infty] {}-8$, so the only problem is with the middle summand, but in fact:

$\frac{2x^2}{e^{x^2\slash 4}}\,\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}\xrigh tarrow [x\to \infty] {} 0$ (Applying L'Hospital to the first factor we get zero, whereas the second one converges to 1) , so the whole thing's bounded.

Tonio

4. Originally Posted by redsoxfan325
$g(x)=\left\{\begin{array}{lr}x^3e^{-x^2/4}\sin(4x^{-2}):&x\neq0\\0:&x=0\end{array}\right\}$

$\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=x^2e^{-x^2/4}\sin(4x^{-2})=\frac{4e^{-x^2/4}\sin(4x^{-2})}{4x^{-2}}$

So $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{4e^{-x^2/4}\sin(4x^{-2})}{4x^{-2}}$

Do you know where to go from here?

See Tonio's post below for proving $g'$ is bounded.
I just take the limit as x -> 0 and from that can determine what f'(0) = .. ? Right? Not confident in that, lol.

5. Originally Posted by katielaw
I just take the limit as x -> 0 and from that can determine what f'(0) = .. ? Right? Not confident in that, lol.
Sorry, I actually mislead you the way I wrote that. I wrote $\frac{\sin(4x^{-2})}{4x^{-2}}$ thinking that it would fall under the $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ category, but it doesn't because $\lim_{x\to0}4x^{-2}\neq0$.

So the way you should actually do the limit is to use the squeeze theorem.

$\lim_{x\to0}-x^2e^{-x^2/4}\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq\lim_{x\to0}x^2e^{-x^2/4}$

Thus, $0\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq0$, so the limit is zero.

Sorry for any confusion I may have caused.

6. Originally Posted by redsoxfan325
Sorry, I actually mislead you the way I wrote that. I wrote $\frac{\sin(4x^{-2})}{4x^{-2}}$ thinking that it would fall under the $\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ category, but it doesn't because $\lim_{x\to0}4x^{-2}\neq0$.

So the way you should actually do the limit is to use the squeeze theorem.

$\lim_{x\to0}-x^2e^{-x^2/4}\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq\lim_{x\to0}x^2e^{-x^2/4}$

Thus, $0\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq0$, so the limit is zero.

Sorry for any confusion I may have caused.
That's more than OK. Thank you so much for your help.