Let f(x)=3 sin ^{4} x+10 sin ^{3} x+6 sin ^{2 | vedclass

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(C) Given $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$.Differentiating with respect to $x$:$f'(x) = 12 \sin^3 x \cos x + 30 \sin^2 x \cos x + 1

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decreasing in $\left(-\frac{\pi}{6}, 0\right)$

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(C) Given $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$.Differentiating with respect to $x$:$f'(x) = 12 \sin^3 x \cos x + 30 \sin^2 x \cos x + 12 \sin x \cos x$$f'(x) = 6 \sin x \cos x (2 \sin^2 x + 5 \sin x + 2)$Factoring the quadratic expression:$f'(x) = 6 \sin x \cos x (2 \sin x + 1)(\sin x + 2)$For $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$,we have:$1. \cos x > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.$2. (\sin x + 2) > 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.$3. (2 \sin x + 1) \ge 0$ for all $x \in \left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$.Thus,the sign of $f'(x)$ depends on $\sin x$:- If $x \in \left(-\frac{\pi}{6}, 0\right)$,then $\sin x - If $x \in \left(0, \frac{\pi}{2}\right)$,then $\sin x > 0$,so $f'(x) > 0$. Thus,$f$ is increasing in $\left(0, \frac{\pi}{2}\right)$.

Let $f(x)=3 \sin ^{4} x+10 \sin ^{3} x+6 \sin ^{2} x-3$,where $x \in\left[-\frac{\pi}{6}, \frac{\pi}{2}\right]$. Then,$f$ is $.....$

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