Question

Find the greatest and the least values of the following function:

$f\left(x\right)=\cos 3x-15\cos x+8$ where $xϵ[\frac{\pi }{3},\frac{3\pi }{2}]$

Answer 1

$f\left(x\right)=\cos 3x-15\cos x+8$ where $xϵ[\frac{\pi }{3},\frac{3\pi }{2}]$

$f^{\prime }\left(x\right)=-3\sin 3x+15\sin x=0$

$\Rightarrow \sin 3x=5\sin x$

$\Rightarrow 3\sin x-4{\sin }^{3}x=5\sin x$

$\Rightarrow -4{\sin }^{3}x=2\sin x$

$\Rightarrow 2\sin x+4{\sin }^{3}x=0$

$\Rightarrow 2\sin x[1+2{\sin }^{2}x]=0$

Now, $\sin x=0$ and ${\sin }^{2}x=\frac{-1}{2}$ Not possible

$x=\pi$ is the only choice because

$xϵ[\frac{\pi }{2},\frac{3\pi }{2}]$

$f^{\prime \prime }\left(x\right)=-9\cos 3x+15\cos x$

$f^{\prime \prime }\left(\pi \right)=-6<0$, therefore $x=\pi$ is the point of maxima.

$f\left(\frac{\pi }{2}\right)=\cos \frac{3\pi }{2}-15\cos \frac{\pi }{2}+8=8$

$f\left(\pi \right)=\cos 3\pi -15\cos \pi +8=-1+15+8=22$

$f\left(\frac{3\pi }{2}\right)=\cos \frac{9\pi }{2}-15\cos \frac{3\pi }{2}+8=8$

$x=\pi ,f\left(x\right)=22$ is the greatest value

$x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2},f\left(x\right)=8$ is the least value.