Minimum of a trigonometric function involving absolute value

Question

Given $f(x) = | \sin( | x | ) | $, I am told to found the local and global minimum and maximum of $f$ (If they exist).

Simply from sketching the graph of $f$ I get that the function maximizes periodically for every $x = k \frac{\pi}{2} $ with $k$ being a whole number, giving that there's no global maximum but infinite local maxima equal to $1$.

Following the same logic, I thought the function would minimize with period $\pi$ to give local minima of $0$. However, when checking for the answer, it appears the function has no global nor local minima.

Answer 1

Your general reasoning is correct. The global minimum is $0$, and it is attained in points of the form $x = k \pi$. The global maximum is 1, and it is attained in points of the form $x= \frac{\pi}{2}+ k \pi$. Obviously all these global maxima/minima are also local maxima/minima.

I find it very strange that the solution says differently... Are you certain of the expression for $f$?

Answer 2

Good insights to start off about the periodicity of the sine function.

Let's think about this in parts - dissecting the function from the outside inward. Remember the definition of absolute value: $$f(x) = | g(x) | = \left\{ \begin{array}{ll} g(x) & g(x) \geq 0 \\ -g(x) & g(x) < 0 \end{array} \right.$$ $$ g(x)=\sin(|x|)=\sin\left( \begin{array}{ll} x & x \geq 0 \\ -x & x < 0 \end{array}\right)$$

By definition, the minima cannot be lower than 0. That does not make it the minima of the function, but it is a good check for the result we get.

So what is the maxima/minima of $\sin(x)$? $$-1 \leq \sin(x)\leq 1$$ So what is the minima applying the absolute value? $$ 0 \leq |\sin(x)| \leq 1 \forall x$$ Those are the maxima/minima of the function, regardless of the $x$ input. Let's solve for $x$: $$\sin x = 0 \rightarrow x = ... ,-2\pi,\pi,0,\pi,2\pi, ...$$ $$\sin x = 1 \rightarrow x = ... ,{-3\pi \over 2},{-\pi \over 2},{\pi \over 2},{3\pi \over 2},...$$ Notice all the negative values, those don't apply since we apply the absolute value to $x$. $$x_{min}=0,\pi,2\pi,...$$ $$x_{max}={\pi \over 2},{3\pi \over 2},...$$ There are no local maxima/minima which are not the global maxima/minima.

Answer 3

By graphing you function it seems that $1$ and $0$ are the maximum and minimum values for the result, Why ?
The answer is: it's well known that the trigonometric functions are commonly used for expressing ratios between sides of right angled triangle, between$(-1,1)$ each $\frac \pi 2$, but by using the absolute value the result can only be positive so the result goes between $0$ and $1$.

So, the function has a unique global maximum of value $1$ at $x=k \frac \pi 2, k \in \mathbb Z$ and a unique global minimum of value $0$ at $x=m, m \in \mathbb Z$.