Which of the following functions is decreasin | vedclass

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Question

(C) To determine which function is decreasing in the interval $I = \left(0, \frac{\pi}{8}\right)$,we examine the derivative of each function:$A) f(x)

Vedclass · Accepted Answer

$\cos 4x$

Vedclass · Answer

(C) To determine which function is decreasing in the interval $I = \left(0, \frac{\pi}{8}ight)$,we examine the derivative of each function:$A) f(x) = 	an 4x \implies f'(x) = 4 \sec^2 4x$. Since $\sec^2 4x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.$B) f(x) = \sin x \implies f'(x) = \cos x$. Since $\cos x > 0$ for all $x \in \left(0, \frac{\pi}{8}ight)$,$f'(x) > 0$,so the function is increasing.$C) f(x) = \cos 4x \implies f'(x) = -4 \sin 4x$. For $x \in \left(0, \frac{\pi}{8}ight)$,$4x \in (0, \frac{\pi}{2})$. In this interval,$\sin 4x > 0$,so $f'(x) = -4 \sin 4x $D) f(x) = -\cos x \implies f'(x) = \sin x$. Since $\sin x > 0$ for all $x \in I$,$f'(x) > 0$,so the function is increasing.Therefore,the correct option is $C$.

Which of the following functions is decreasing in the interval $\left(0, \frac{\pi}{8}\right)$?

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