Let $f(x)=3(x-2)^{\frac{2}{3}}-(x-2),~0\leq x\leq 20$ Let $x_0$ and $y_0$ be the points of the global minima and maxima, respectively, of $f(.)$ in the interval $[0,20]$. Evaluate $f(x_0)+f(y_0)$
Note that $f'(x)=2(x-2)^{-\frac{1}{3}}-1=0$ $=>x=10$ and $f''(10)=-\frac{2}{3}(10-2)^{-\frac{4}{3}}=-10.67<0$ So,at $x=10,~f(x)$ is max.
But Note that $f(10)=4$ but $f(0)=6.76$ I think I make some mistake but I can't find it out .Please solve the problem