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Please I need a hand in solving this problem:
These 3 functions' values at $0$ are all $0$ and for $x\ne0$, $$f(x)=x^4\sin\frac{1}{x}, \, g(x)=x^4\left(2+\sin\frac{1}{x}\right), \, h(x)=x^4\left(-2+\sin\frac{1}{x}\right)$$
b- Show that $f$ has neither a local maximum nor a local minimum at $0$, $g$ has a local minimum, and $h$ has a local maximum.
The derivatives of these functions change sign infinitely often on both sides of $0$. I couldn't use the 1st nor the 2nd derivative test.
The behavior of these functions is easily handled without using derivatives. Note that for any given open interval $I$ containing $0$ we can find positive integer $ n$ such that points $c, d$ given by $$c=\frac{2}{(4n+1)\pi},d=\frac{2}{(4n+3)\pi}$$ lie in $I$ and $f(c) >0=f(0)>f(d)$. So $0$ is neither a local maximum nor a local minimum of $f$.
Functions $g, h$ are easier to handle. For $g$ note that $(2+\sin(1/x))>0$ so that $g(x)$ is positive for all $x$ except $x=0$ and similarly $h(x) $ is always negative except at $x=0$. The desired conclusion follows immediately.
HINT: we have for 1) $$f'(x)=4x^3\sin\left(\frac{1}{x}\right)+x^4\cos\left(\frac{1}{x}\right)\cdot \left(-\frac{1}{x^2}\right)$$ and for 2)$$g'(x)=4x^3\left(2+\sin\left(\frac{1}{x}\right)\right)+x^4\cdot\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$ and 3)$$h'(x)=4x^3\left(-2+\sin\left(\frac{1}{x}\right)\right)+x^4\cos\left(\frac{1}{x}\right)\left(-\frac{1}{x^2}\right)$$ additionally use that $$|f(x)|\le x^4$$ and for $$f(x)$$ we have a sign Change from minus to plus $g(x)$ has a sign change from plus to plus $h(x)$ has a sign change for minus to minus.....
