# Exercise 66 of chapter 11 in Spivak calculus

Exercise 66 of chapter 11 in Spivak calculus, asks to show that (a) $f$ has a local minimum at $0$ and (b) prove that $f'(0)=f''(0)=0$ for the function $$f(x)=\begin{cases} x^4\sin^2\left(\frac{1}{x}\right)&\text{for x\not=0}\\ 0&\text{for x=0} \end{cases}$$ I have done these two parts and I am ok with it. However at the end of this question, there is a comment that I do not understand a part of this comment (I have written the part that I do not understand in bold).

This function thus provides another example to show that theorem 6 can not be improved. It also illustrates a subtlety about maxima and minima that often goes unnoticed; a function may not be increasing in any interval to the right of a local minimum point, nor decreasing in any interval to the left.

Could someone please clarify this for me? and here is theorem 6.

Theorem 6. Suppose $f''(a)$ exists. If $f$ has a local minimum at $a$, then $f''(a)\geq0$; if $f$ has a local maximum at $a$, then $f''(a)\leq0$.

• How does Spivak define the term "local minimum"? – user142299 May 21 '14 at 3:52
• Also, how did you show that it is a minimum? Presumably you had to calculate the derivatives first, so why are two of them asked for in the second part of the question? – user142299 May 21 '14 at 4:03
• For the first part, you do not really need derivatives; $f(0)=0$ and so $f(x)\geq0$ for all $x\in\mathbb{R}$. Thus $0$ is a minimum for all $x\in\mathbb{R}$. – user39193 May 21 '14 at 4:06
• Show that $\sin^2(1/x)$ is not monotonic in any interval $(-\delta,\delta)$ by showing the derivative changes sign at infinitely many points. – RRL May 21 '14 at 4:12
• @NotNotLogical Good point, but $f(x) = c$ isn't quite as good of an example because it is both increasing and decreasing on both sides of the minima, just not strictly increasing or decreasing. This function $f$ is not even increasing or decreasing at all. – 6005 May 21 '14 at 5:36

## 1 Answer

It may help you to look at the graph of $f$: At any value of $x = \frac{1}{k\pi}$ for some $k$, the graph is zero. So it is clear that despite having a local minimum at $0$ (defining "local minimum" to include the non-strict inequality $f(0) \le f(x)$), $f$ is not increasing or decreasing in any interval to the left or right. Here I'm using "increasing" to be non-strict, so it means $a < b$ implies $f(a) \le f(b)$. Similarly for decreasing.