# differentiable, f' not continuous

• Mar 19th 2009, 02:46 PM
chloe561
differentiable, f' not continuous
Let $\displaystyle f(x)=x^2 \sin(\frac{1}{x})$ for $\displaystyle x \not = 0$ and $\displaystyle f(0)=0.$

a) Prove that $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.
b) Prove that $\displaystyle f'(x)$ is not continuous on $\displaystyle \mathbb{R}$.

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

Attempt:
a) Clearly $\displaystyle f(x)$ is differentiable at $\displaystyle x \not = 0$. [How do I show this?] In fact, $\displaystyle f'(x)=2x \sin(x) - \cos(\frac{1}{x})$ [valid for $\displaystyle x \not =0$]. At $\displaystyle x=0$, $\displaystyle f'(0)=\lim_{x \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim_{h \rightarrow 0} h \sin(\frac{1}{h})=0$. Hence, $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.

b) $\displaystyle f'(x)=2x \sin(x) - \cos(\frac{1}{x})$. Let $\displaystyle p(x)=2x \sin(x)$. Then $\displaystyle p(0)=0$ because $\displaystyle p(0)=2 \cdot \lim_{x \rightarrow 0} x \sin(\frac{1}{x})=0$. So $\displaystyle p(0)=0$. So by FTC $\displaystyle \exists P(x)$ s.t. $\displaystyle P'(x)=p(x)$. So $\displaystyle f'(x)=P'(x)-cos(\frac{1}{x})$. But then $\displaystyle \cos(\frac{1}{x})$ is not a continuous function. But by FTC, $\displaystyle P(x)$ is differentiable. Then $\displaystyle P(x)$, $\displaystyle f(x)$ are differentiable functions whose derivatives are not continuous.

Thanks.
• Mar 19th 2009, 04:11 PM
Krizalid
$\displaystyle f'(x)\to\infty$ as $\displaystyle x\to0,$ thus $\displaystyle f'(x)$ is not continuous at $\displaystyle x=0.$ $\displaystyle f$ is of course, differentiable but not of class 1.
• Mar 21st 2009, 06:08 AM
Laurent
Quote:

Originally Posted by chloe561
Let $\displaystyle f(x)=x^2 \sin(\frac{1}{x})$ for $\displaystyle x \not = 0$ and $\displaystyle f(0)=0.$

a) Prove that $\displaystyle f(x)$ is differentiable on $\displaystyle \mathbb{R}$.
b) Prove that $\displaystyle f'(x)$ is not continuous on $\displaystyle \mathbb{R}$.

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

For part a), to justify that $\displaystyle f$ is differentiable on $\displaystyle \mathbb{R}\setminus\{0\}$, you can say that it is obtained by product and composition of the following differentiable functions: $\displaystyle x\mapsto x^2$ on $\displaystyle \mathbb{R}$, $\displaystyle \sin$ on $\displaystyle \mathbb{R}$ and $\displaystyle x\mapsto\frac{1}{x}$ on $\displaystyle \mathbb{R}^*$. The computation at 0 is fine (well, it would be better not to write $\displaystyle f'(0)$ when you don't know that it exists yet)

For part b), you don't need to refer to FTC : just say that $\displaystyle x\sin\frac{1}{x}$ tends to 0, while $\displaystyle \cos\frac{1}{x}$ has no limit as $\displaystyle x\to0^+$, so that $\displaystyle f'(x)$ has no limit as $\displaystyle x\to0^+$. Therefore, it can't be continuous (whatever $\displaystyle f'(0)$ is).

Quote:

Originally Posted by Krizalid
$\displaystyle f'(x)\to\infty$ as $\displaystyle x\to0,$ thus $\displaystyle f'(x)$ is not continuous at $\displaystyle x=0.$ $\displaystyle f$ is of course, differentiable but not of class 1.

No, $\displaystyle f'(x)$ is bounded near 0 (and it has no limit when $\displaystyle x\to0^+$).