Question
Show that the function given by f (x) = sin x is neither increasing nor decreasing in (0, $\pi$)

Answer

The function is f (x) = sin x
Then, $f^\prime$(x) = cos x
Since for each x $\in$ $\left(0, \frac{\pi}{2}\right)$, cos x > 0, we have $f^\prime(x)$ >0
Therefore, $f$ is strictly increasing in$\left(0, \frac{\pi}{2}\right)$……(1)
Now, The function is f (x) = sin x
Then, $f^\prime(x)=cosx$
Since, for each  $x\in\left(\frac{\pi}{2}, \pi\right)$, cos x < 0, we have $f^\prime(x)$ < 0
Therefore, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$……(2)
From (1) and (2),
It is clear that f is neither increasing nor decreasing in (0, $\pi$).

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