Differentiating composite function involving sin(1/x)

Question

For $$f(x) = \begin{cases} x+2x^2\sin(1/x), & \text{if $x \ne$ 0} \\ 0, & \text{if $x=0$ } \end{cases}$$ I need to show that $f'(0)>0$ but also that $f$ is not increasing in any interval around 0. I don't know how to find the derivative of $f$ at 0 because $f'(x)$ is undefined at that point. As to the second part of the problem (showing that $f$ is not increasing in any interval around 0) I simply do not know where to start in order to prove this. All help is appreciated.

Answer 1

To compute $f'(0),$ go back to the definition of derivative:

$$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}.$$

You should be able to figure out this limit. (Where the formula refers to $f(h),$ $h$ is non-zero, so we're in the first case of your definition.)

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To show that $f$ isn't increasing in any interval around $0,$ you could try to find arbitrarily small positive $x$ such that $f(x)\lt 0.$

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By the way, you claimed that $f'(x)$ is undefined at $x=0.$ That's not true. (What you probably mean is that the formula that you get when you work out $\,f'(x)$ for $x\ne 0$ is not defined at $x=0,$ but that's not the same as computing $\,f'(0).)$