Question

The number of points of local extremum of $f(x)=|\sin x|$ over the interval (0,2$\pi$) is (are)

Answer 1

3

Answer 2

To find the number of points of local extremum for the function $f(x)=|\sin x|$ over the interval $(0, 2\pi)$, we need to analyze the function's behavior.

The function $f(x) = |\sin x|$ can be defined piecewise in the given interval:

1. For $x \in (0, \pi)$, $\sin x > 0$, so $f(x) = \sin x$.
2. For $x = \pi$, $\sin x = 0$, so $f(\pi) = 0$.
3. For $x \in (\pi, 2\pi)$, $\sin x < 0$, so $f(x) = -\sin x$.

We will examine the derivative of $f(x)$ in each interval and the behavior at the point where the definition changes ($x=\pi$).

1. Analysis for $x \in (0, \pi)$:

$f(x) = \sin x$
$f'(x) = \cos x$

To find critical points, set $f'(x) = 0$:
$\cos x = 0 \implies x = \frac{\pi}{2}$ in $(0, \pi)$.

Let's check the sign of $f'(x)$ around $x = \frac{\pi}{2}$:

• For $x \in (0, \frac{\pi}{2})$, $\cos x > 0$, so $f(x)$ is increasing.
• For $x \in (\frac{\pi}{2}, \pi)$, $\cos x < 0$, so $f(x)$ is decreasing.

Since $f'(x)$ changes from positive to negative at $x = \frac{\pi}{2}$, $x = \frac{\pi}{2}$ is a point of local maximum.
$f(\frac{\pi}{2}) = |\sin(\frac{\pi}{2})| = 1$.

2. Analysis for $x \in (\pi, 2\pi)$:

$f(x) = -\sin x$
$f'(x) = -\cos x$

To find critical points, set $f'(x) = 0$:
$-\cos x = 0 \implies \cos x = 0 \implies x = \frac{3\pi}{2}$ in $(\pi, 2\pi)$.

Let's check the sign of $f'(x)$ around $x = \frac{3\pi}{2}$:

• For $x \in (\pi, \frac{3\pi}{2})$, $\cos x < 0$, so $-\cos x > 0$. Thus $f'(x) > 0$, and $f(x)$ is increasing.
• For $x \in (\frac{3\pi}{2}, 2\pi)$, $\cos x > 0$, so $-\cos x < 0$. Thus $f'(x) < 0$, and $f(x)$ is decreasing.

Since $f'(x)$ changes from positive to negative at $x = \frac{3\pi}{2}$, $x = \frac{3\pi}{2}$ is a point of local maximum.
$f(\frac{3\pi}{2}) = |\sin(\frac{3\pi}{2})| = |-1| = 1$.

3. Analysis at the junction point $x = \pi$:

First, check continuity at $x = \pi$:
$\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^-} \sin x = \sin \pi = 0$.
$\lim_{x \to \pi^+} f(x) = \lim_{x \to \pi^+} (-\sin x) = -\sin \pi = 0$.
$f(\pi) = |\sin \pi| = 0$.

Since the limits from both sides and the function value are equal, $f(x)$ is continuous at $x = \pi$.

Next, check differentiability and the behavior around $x = \pi$:
Left-hand derivative: $f'(\pi^-) = \lim_{x \to \pi^-} \cos x = \cos \pi = -1$.
Right-hand derivative: $f'(\pi^+) = \lim_{x \to \pi^+} (-\cos x) = -\cos \pi = -(-1) = 1$.

Since $f'(\pi^-) \neq f'(\pi^+)$, the derivative $f'(x)$ does not exist at $x = \pi$. This indicates a sharp corner (cusp).

Let's analyze the monotonicity around $x = \pi$:

• For $x < \pi$ (specifically, in an interval like $(\frac{\pi}{2}, \pi)$), $f'(x) = \cos x < 0$. So, $f(x)$ is decreasing.
• For $x > \pi$ (specifically, in an interval like $(\pi, \frac{3\pi}{2})$), $f'(x) = -\cos x > 0$. So, $f(x)$ is increasing.

Since $f(x)$ is continuous at $x = \pi$ and changes from decreasing to increasing at $x = \pi$, $x = \pi$ is a point of local minimum.
$f(\pi) = |\sin \pi| = 0$.

Conclusion:

The points of local extremum in the interval $(0, 2\pi)$ are:

• $x = \frac{\pi}{2}$ (local maximum)
• $x = \pi$ (local minimum)
• $x = \frac{3\pi}{2}$ (local maximum)

There are 3 points of local extremum.