$\text { Find all critical numbers of the function } f ( x ) = \sin ^ { 2 } 6 x + \cos 6 x , 0

Question

$\text { Find all critical numbers of the function } f ( x ) = \sin ^ { 2 } 6 x + \cos 6 x , 0 < x < \frac { \pi } { 3 } \text {. }$ 
A) $\frac { \pi } { 18 } , \frac { \pi } { 12 } , \frac { \pi } { 9 } , \frac { \pi } { 4 }$
B) $\frac { \pi } { 18 } , \frac { \pi } { 6 } , \frac { 5 \pi } { 18 }$
C) $\frac { \pi } { 36 } , \frac { \pi } { 6 } , \frac { 2 \pi } { 9 }$
D) $\frac { \pi } { 24 } , \frac { \pi } { 6 } , \frac { \pi } { 2 }$
E) $\frac { \pi } { 6 } , \frac { 5 \pi } { 24 } , \frac { 7 \pi } { 24 }$

Quizplus · Accepted Answer

To find the critical numbers of $f(x) = \sin^2 6x + \cos 6x$, we first find its derivative and set it equal to zero. The derivative is $f'(x) = 2\sin 6x \cdot 6\cos 6x - 6\sin 6x = 12\sin 6x \cos 6x - 6\sin 6x$. Setting $f'(x) = 0$ gives $12\sin 6x \cos 6x - 6\sin 6x = 0$, which simplifies to $\sin 6x(2\cos 6x - 1) = 0$. This equation has solutions when $\sin 6x = 0$ and when $2\cos 6x - 1 = 0$. Solving within the given interval $0 < x < \frac{\pi}{3}$ leads to the critical points $\frac{\pi}{18}, \frac{\pi}{6}, \frac{5\pi}{18}$, which matches option B.

Find all critical numbers of the function f(x)=sin⁡26x+cos⁡6x,0<x<π3. \text { Find all critical numbers of the function } f ( x ) = \sin ^ { 2 } 6 x + \cos 6 x , 0 < x < \frac { \pi } { 3 } \text {. } Find all critical numbers of the function f(x)=sin26x+cos6x,0<x<3π​.

$\text { Find all critical numbers of the function } f ( x ) = \sin ^ { 2 } 6 x + \cos 6 x , 0 < x < \frac { \pi } { 3 } \text {. }$