Question

Show that the function given by $f(x) = sin x$ is (a) strictly increasing in $(0,\frac{π}{2})$ (b) strictly decreasing in $(\frac{π}{2},π)$ (c) neither increasing nor decreasing in $(0, π)$

Answer 1

The given function is $f(x) = sin x$.
$∴ f'(x)=cos x$
(a) Since for each $x∈(0,\frac{π}{2}),cosx>0$, we have $f'(x)>0$
Hence, $f$ is strictly increasing in $(0,\frac{π}{2})$
(b) Since for each $x∈(\frac{π}{2}),cos x<0$, we have $f'(x)<0$
Hence, $f$ is strictly decreasing in $(\frac{π}{2π})$.
(c) From the results obtained in (a) and (b), it is clear that $f$ is neither increasing nor decreasing in $(0, π)$.