use the derivative fto determine the local minima and maxima of f and the intervals of increase and…

use the derivative fto determine the local minima and maxima of f and the intervals of increase and decrease. sketch a possible graph of f (f is not unique). f(x)=10 sin 2x on -2π,2π. the local minimum/minima is/are at x = -π,0,π. (use a comma to separate answers as needed. type an exact answer, using π as needed.) the local maximum/maxima is/are at x = . (use a comma to separate answers as needed. type an exact answer, using π as needed.)

use the derivative fto determine the local minima and maxima of f and the intervals of increase and decrease. sketch a possible graph of f (f is not unique). f(x)=10 sin 2x on -2π,2π. the local minimum/minima is/are at x = -π,0,π. (use a comma to separate answers as needed. type an exact answer, using π as needed.) the local maximum/maxima is/are at x = . (use a comma to separate answers as needed. type an exact answer, using π as needed.)

Answer

Explanation:

Step1: Find critical points

Set $f^{\prime}(x)=0$. So, $10\sin2x = 0$. Then $\sin2x=0$. We know that $\sin\theta = 0$ when $\theta = k\pi$, $k\in\mathbb{Z}$. So, $2x=k\pi$, and $x=\frac{k\pi}{2}$, where $k\in\mathbb{Z}$. In the interval $[- 2\pi,2\pi]$, when $k=-4,x = - 2\pi$; when $k=-3,x=-\frac{3\pi}{2}$; when $k = - 2,x=-\pi$; when $k=-1,x =-\frac{\pi}{2}$; when $k = 0,x = 0$; when $k = 1,x=\frac{\pi}{2}$; when $k = 2,x=\pi$; when $k = 3,x=\frac{3\pi}{2}$; when $k = 4,x = 2\pi$.

Step2: Use the first - derivative test

We divide the interval $[-2\pi,2\pi]$ into sub - intervals using the critical points. Consider the sign of $f^{\prime}(x)=10\sin2x$ in each sub - interval. Let's take test points: For the interval $(-2\pi,-\frac{3\pi}{2})$, take $x =-\frac{7\pi}{4}$, then $f^{\prime}(-\frac{7\pi}{4})=10\sin(-\frac{7\pi}{2})=10>0$. For the interval $(-\frac{3\pi}{2},-\pi)$, take $x =-\frac{5\pi}{4}$, then $f^{\prime}(-\frac{5\pi}{4})=10\sin(-\frac{5\pi}{2})=- 10<0$. For the interval $(-\pi,-\frac{\pi}{2})$, take $x =-\frac{3\pi}{4}$, then $f^{\prime}(-\frac{3\pi}{4})=10\sin(-\frac{3\pi}{2})=10>0$. For the interval $(-\frac{\pi}{2},0)$, take $x =-\frac{\pi}{4}$, then $f^{\prime}(-\frac{\pi}{4})=10\sin(-\frac{\pi}{2})=-10<0$. For the interval $(0,\frac{\pi}{2})$, take $x=\frac{\pi}{4}$, then $f^{\prime}(\frac{\pi}{4})=10\sin(\frac{\pi}{2})=10>0$. For the interval $(\frac{\pi}{2},\pi)$, take $x=\frac{3\pi}{4}$, then $f^{\prime}(\frac{3\pi}{4})=10\sin(\frac{3\pi}{2})=-10<0$. For the interval $(\pi,\frac{3\pi}{2})$, take $x=\frac{5\pi}{4}$, then $f^{\prime}(\frac{5\pi}{4})=10\sin(\frac{5\pi}{2})=10>0$. For the interval $(\frac{3\pi}{2},2\pi)$, take $x=\frac{7\pi}{4}$, then $f^{\prime}(\frac{7\pi}{4})=10\sin(\frac{7\pi}{2})=-10<0$. A function $y = f(x)$ has a local maximum at $x = c$ if $f^{\prime}(x)$ changes sign from positive to negative at $x = c$, and a local minimum at $x = c$ if $f^{\prime}(x)$ changes sign from negative to positive at $x = c$. The local maxima occur at $x=-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$.

Answer:

$-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}$