The number of critical points of the function f(x)= array|c|c| |( x)/(x)|, & x 0 1, & x=0 …

Question

The number of critical points of the function

$f(x)= \begin{array}{|c|c|} |\frac{\sin x}{x}|, & x
e 0 \ 1, & x=0 \end{array}$ in the interval $(-2\pi ,2\pi )$ is equal to :

Accepted Answer

First check continuity at $x=0$:

$\underset{x	o 0}{lim}|\frac{\sin x}{x}|=1=f(0)	o f(x)$ is continuous

For $x
e 0$, write $g(x)=\frac{\sin x}{x}$. Then

Now $\frac{d}{dx}(\frac{\sin x}{x})=\frac{x\cos x-\sin x}{{x}^{2}}$

To find critical points, we look for points in $(-2\pi ,2\pi )$ where either ${f}^{'}(x)=0$ or ${f}^{'}(x)$ does not exist.

At points where $\frac{\sin x}{x}
e 0$, the absolute value does not create a corner, so we can use ${g}^{'}(x)=0$ to get stationary points:

$\Rightarrow {f}^{'}(x)=0\Rightarrow 	an x=x$

The equation $	an x=x$ has 3 solutions in $(-2\pi ,2\pi )$ (one is $x=0$, and one each in $(\pi ,\frac{3\pi }{2})$ and $(-\frac{3\pi }{2},-\pi )$):

Next, ${f}^{'}(x)$ does not exist at the zeros of $|\frac{\sin x}{x}|$, because $|\, \cdot \, |$ creates a sharp corner when the inside changes sign. Here $\sin x=0$ gives $x=\pm \pi $ (and $\pm 2\pi $ etc.). In the open interval $(-2\pi ,2\pi )$, only $x=-\pi $ and $x=\pi $ lie inside.

So we have $3$ critical points from $	an x=x$, and $2$ more critical points where ${f}^{'}(x)$ does not exist at $x=-\pi ,\pi $.

total 5 points

Solutions · 1