Question
Prove that the function given by f(x) = cos x is neither increasing nor decreasing in (0, 2$\pi$).
✓
Answer
Note that $f ′(x) = – sin x$
Clearly, for $$$(0, \pi)$, we have
$f ′(x) = – sin x<0$
So, the function is decreasing in $(0, \pi)$.
Now, for $x \in (\pi, 2\pi)$ we have
$f ′(x) = – sin x>0$
Therefore, $f(x)$ is increasing in $(\pi, 2\pi)$
Hence, f is neither increasing nor decreasing in (0, 2$\pi$).
Clearly, for $$$(0, \pi)$, we have
$f ′(x) = – sin x<0$
So, the function is decreasing in $(0, \pi)$.
Now, for $x \in (\pi, 2\pi)$ we have
$f ′(x) = – sin x>0$
Therefore, $f(x)$ is increasing in $(\pi, 2\pi)$
Hence, f is neither increasing nor decreasing in (0, 2$\pi$).
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