The absolute minimum value of f(x)=2 x in [0, 3 2 ] is - Vidyadip

Question

The absolute minimum value of $f(x)=2 \sin x$ in $\left[0, \frac{3 \pi}{2}\right]$ is

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Answer

(a) : Here, $f(x)=2 \sin x$
$\Rightarrow f^{\prime}(x)=2 \cos x$
Putting $f^{\prime}(x)=0 \Rightarrow 2 \cos x=0 \Rightarrow \cos x=0$
$\Rightarrow x=\frac{\pi}{2}, \frac{3 \pi}{2}$
$\therefore \quad \frac{\pi}{2}, \frac{3 \pi}{2}$ are the critical points.
At $x=0, f(x)=2 \sin (0)=0$
At $x=\frac{\pi}{2}, f(x)=2 \times 1=2$
At $x=\frac{3 \pi}{2}, f(x)=2 \sin \left(\pi+\frac{\pi}{2}\right)=-2 \sin \frac{\pi}{2}=-2$
Hence, absolute minimum value of $f(x)$ is -2 .

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