Question

Find the absolute maximum and minimum values of the function f given by $f\left(x\right)=co{s}^{2}x+sinx,xϵ[0,\pi ]$

Answer 1

Given, $f\left(x\right)=co{s}^{2}x+sinx,xϵ[0,\pi ]$
Now, $f^{\prime }\left(x\right)=2cosx(-sinx)+cosx=-2sinxcosx+cosx$
For maximum or minimum put $f^{\prime }\left(x\right)=0\Rightarrow -2sinxcosx+cosx=0$
$\Rightarrow cosx(-2sinx+1)=0\Rightarrow cosx=0orsinx=\frac{1}{2}\Rightarrow x=\frac{\pi }{6},\frac{\pi }{2}$
For absolute maximum and absolute minimum, we have to evaluate
$f\left(0\right),f\left(\frac{\pi }{6}\right),f\left(\frac{\pi }{2}\right),f\left(\pi \right)$
At $x=0,$ $f\left(0\right)=co{s}^{2}0+sin0=1^2+0=1$
At $x=\frac{\pi }{6},$ $f\left(\frac{\pi }{6}\right)=co{s}^2\left(\frac{\pi }{6}\right)+sin\frac{\pi }{6}=(\frac{\sqrt{3}}{2}{)}^2+\frac{1}{2}=\frac{5}{4}=1.25$
At $x=\frac{\pi }{2},$ $f\left(\frac{\pi }{2}\right)=co{s}^2\left(\frac{\pi }{2}\right)+sin\frac{\pi }{2}=0^2+1=1$
At $x=\pi ,$ $f\left(\pi \right)=co{s}^2\pi +sin\pi =(-1{)}^2+0=1$
Hence, the absolute maximum value of f is 1.25 occuring at $x=\frac{\pi }{6}$ and the absolute minimum value of f is 1 occuring at x = 0, $\frac{\pi }{2}and\pi .$
Note If close interval is given, to determine global maximum (minimum), check the value at all critical points as well as end points of a given interval.