Question

Find the greatest and the least values of the functions $f\left(x\right)=5(1+\sin x)\cos x+3x$ in the interval $[0,\frac{\pi }{2}].$

Answer 1

$f\left(x\right)=5(1+\sin x)\cos x+3x$ in the interval $[0,\frac{\pi }{2}]$

$f\left(x\right)=5\cos x+\frac{5}{2}\sin 2x+3x$

$f^{\prime }\left(x\right)=-5\sin x+\frac{5}{2}\times 2\cos 2x+3$

$=-5\sin x+5[1-2{\sin }^{2}x]+3$

$=-5\sin x+5-10{\sin }^{2}x+3$

$=-10{\sin }^{2}x-5\sin x+8=0$

$\Rightarrow 10{\sin }^{2}x+5\sin x-8=0$

$\sin x=\frac{-25\pm \sqrt{25+4\left(8\right)\left(10\right)}}{20}$

We know that $\sqrt{345}=19$

$\therefore$ $\sin x<0$

$\Rightarrow x\text{/}ϵ[0,\frac{\pi }{2}]$

$f\left(0\right)=[5+\sin 0]\cos 0+3\left(0\right)$

$=5[\because \cos 0=1]$

$f\left(\frac{\pi }{2}\right)=[5+\sin \frac{\pi }{2}]\cos \frac{\pi }{2}+3\frac{\pi }{2}$

$=0+\frac{3\pi }{2}=4.7$

$\therefore$ Greatest value at $x=0,f\left(0\right)=5$

and min value at $x=\frac{\pi }{2},f\left(\frac{\pi }{2}\right)=4.7$.