Question

Find the absolute maximum and minimum values of the function \(f\) given by \(f(x)=cos^2x+sinx,x∈[0,π]\)

Answer 1

\(f(x)=cos^2x+sinx\)
\(f'(x)=2cosx(-sinx)+cosx\)
\(=-2sinx\,cosx+cosx\)
Now,\(f(x)=0\)
\(⇒2sinx\,cosx=cosx\)\(⇒cosx(2sinx-1)\)
\(⇒sinx=\frac{1}{2}\, or\, cosx=0\)
\(⇒x=\frac{π}{6},or \frac{π}{2} as x∈[0,π]\)
Now, evaluating the value of \(f\) at critical points \(x=\frac{π}{2}\) and \(x=\frac{π}{6}\) and at the end points of the interval \([0,π]\) (i.e., at \(x = 0\) and \(x = π\)), we have:
\(f(\frac{π}{6})=cos^2\frac{π}{6}+sin\frac{π}{6}\)
\(=(\sqrt{\frac{3}{2}})^2+\frac{1}{2}=\frac{5}{4}\)
\(f(0)=cos^20+sin0=1+0=1\)
\(f(π)=cos^2π+sinπ=(-1)^2+0=1\)
\(f(\frac{π}{2})=cos^2\frac{π}{2}+sin\frac{π}{2}=0+1=1\)
Hence, the absolute maximum value of f is \(\frac{5}{4}\) is occurring at \(x=\frac{π}{6}\) and the absolute minimum value of \(f\) is 1 occurring at \(x=0,π/2\),and \(π\).