Question

Find the intervals in which the function $f$ given by $f\left(x\right)={\sin }^{4}x+{\cos }^{4}x,x\in [0,\pi /2]$ is
$\left(a\right)$ is decreasing
$\left(b\right)$ is increasing
Also, state whether $x=\pi /4$ is a point of local minima or local maxima ?

Answer 1

$f\left(x\right)={\sin }^{4}x+{\cos }^{4}x,x\in [0,\pi /2]$
$f^{\prime }\left(x\right)=4{\sin }^{3}x\cos x-4{\cos }^{3}x\sin x$
$=-4\sin x\cos x({\cos }^{2}x-{\sin }^{2}x)$
$=-2\sin 2x\cdot \cos 2x$
$=-\sin 4x$

Now, $f^{\prime }\left(x\right)=0$ gives $x=0,\pi /4$
The point $x=\pi /4$ divides the interval $[0,\pi /2]$ into two disjoint intervals $[0,\pi /4)$ and $(\pi /4,\pi /2]$

When $0\le x<\pi /4$
$\Rightarrow 0\le 4x<\pi$
Since, $\sin y$ is positive in $0$ to $\pi$.
$\Rightarrow \sin 4x>0$
$\Rightarrow -\sin 4x<0$
$\Rightarrow f^{\prime }\left(x\right)<0$
$\therefore f\left(x\right)$ is decreasing on $[0,\pi /4)$

When $\pi /4<x\le \pi /2$
$\Rightarrow \pi <4x\le 2\pi$
Since, $\sin y$ is negative in $\pi$ to $2\pi$.
$\Rightarrow \sin 4x<0$
$\Rightarrow -\sin 4x>0$
$\Rightarrow f^{\prime }\left(x\right)>0$
$\therefore f\left(x\right)$ is increasing on $(\pi /4,\pi /2]$

$x=\pi /4$
for $x<\pi /4,f^{\prime }\left(x\right)<0$
for $x>\pi /4,f^{\prime }\left(x\right)>0$
$\Rightarrow x=\pi /4$ is local minima.