0
$\begingroup$

Consider the function

$$f(x)=\begin{cases} x^2 \sin{\left ( \frac{1}{x^2} \right )},\ \text{ x} \neq \text{0} \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{x} = \text{0} \end{cases}$$

Here is the graph of $f$

enter image description here

Since $$f'(x)=2x\sin{(1/x^2)}-\frac{2\cos{(1/x^2)}}{x}$$ we can see that $f'$ is not bounded below on $[0,1]$.

Is $f'$ defined at $0$?

Is $f$ is differentiable on $[0,1]$?

$\endgroup$
2
  • 2
    $\begingroup$ What have you tried? As stated this looks like a question with no effort put in. $\endgroup$ May 12 at 17:26
  • $\begingroup$ no f is diffrentiable at x=0 , see my solution $\endgroup$ May 12 at 17:47

2 Answers 2

1
$\begingroup$

Firstly, for any non zero $x$, $f$ is diffrentiable. It is easy to prove. now, I am trying to prove for $x=0$.

$$f^{\prime}(0)=\lim_{x\to 0}\frac{x^2\sin(\frac{1}{x^2})}{x}=\lim_{x\to 0} x\sin(\frac{1}{x^2}).$$

Now, $RHL=\lim_{x\to 0^+} x\sin(\frac{1}{x^2})=0$ and $LHL=\lim_{x\to 0^-} x\sin(\frac{1}{x^2})=0$.

(Since $\sin(\frac{1}{x^2})$ is a bounded function so the limit is $0$ as $x\to 0$.)

Since both limits are equal, $f$ is differentiable at $x=0$.

Basically, $f$ is diffrentiable on $[0,1]$, but $f^{\prime}$ is not continious at $x=0$.

$\endgroup$
0
0
$\begingroup$

$$f(x)=\begin{cases} x^2 \sin{(1/x^2)}, \text{ x} \neq 0 \\ 0,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ x} = 0 \end{cases}$$

$$f'(0)=\lim\limits_{h \to 0} \frac{f(0+h)-f(0)}{h-0}$$

$$=\lim\limits_{h \to 0} \frac{h^2\sin{(1/h^2)}}{h}$$

$$=\lim\limits_{h \to 0} h \sin{(1/h^2)}$$

$$=0$$

Therefore, $f$ is differentiable at $0$.

For $x \neq 0$,

$$f'(x)=2x\sin{(1/x^2)}-\frac{2\cos{(1/x^2)}}{x}$$

$$\lim\limits_{x \to 0^+} f'(x) = -\infty$$ $$\lim\limits_{x \to 0^-} f'(x) = +\infty$$

Hence, $f'$ is not continuous at $0$.

$f'$ is defined at $0$, $f$ is differentiable on $[0,1]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.