Answer to Question #273994 in Calculus for Josa Kyme Palma

Question

Consider the function f(x)=sin x cos x+5 on the interval (0, 2pi).

(a) find the open interval(s) on which the function increasing or decreasing, and

(b) apply the first derivative test to identify the relative extrema.

Answer 1

Let us consider the function $f(x)=\sin x \cos x+5=\frac{1}{2}\sin(2x)+5$ on the interval $(0, 2\pi).$

(a) Let us find the open intervals on which the function increasing or decreasing. Since $f'(x)=\cos(2x),$ and $\cos(2x)=0$ implies $2x=\frac{\pi}2+\pi n,$ that is $x=\frac{\pi}4+\frac{\pi}2 n,$ we conclude that $f'(x)>0$ on the open intervals $(0,\frac{\pi}4),\ (\frac{3\pi}4,\frac{5\pi}4)$, $(\frac{7\pi}4, 2\pi),$ and $f'(x)<0$ on the open intervals $(\frac{\pi}4, \frac{3\pi}4),\ (\frac{5\pi}4,\frac{7\pi}4).$ It follows that the function $f(x)$ increasing on the open intervals $(0,\frac{\pi}4),\ (\frac{3\pi}4,\frac{5\pi}4)$, $(\frac{7\pi}4, 2\pi),$ and decreasing on the open intervals $(\frac{\pi}4, \frac{3\pi}4),\ (\frac{5\pi}4,\frac{7\pi}4).$

(b) Let us identify the relative extrema. It follows that the points $\frac{\pi}4, \ \frac{5\pi}4$ are the points at which the function has the local maximum, and the points $\frac{3\pi}4, \ \frac{7\pi}4$ are the points at which the function has the local minimum.