Finding the maximum value of the function $( x^2-x+1)^{ 1/3}$ on the interval $[0,1]$ While finding the maximum value of the function $( x^2-x+1)^{ 1/3}$ on the closed interval $[0,1]$ the point where the derivative is $0$ is $1/2$ 
and $f(1/2)=(3/4)^{1/3}$. 
Then,  for $f(0)=1$ and for $f(1)=1$... Now here I have a problem, for in my book from both options $0$ and $1$, the answer is only one. Can $0$ also be the answer? Why? 
 A: I find $f(0)=f(1)=-1$, but I also get $f(1/2)=-\left(\frac 54\right)^{1/3}$  Are you sure you have the function right?  As written, the maximum is both at $0$ and $1$  Here is the Alpha plot
A: If $f(x) = (x^2 - x - 1)^{1/3}$, the derivative is
$$
f'(x) = \frac{1}{3}(x^2 - x - 1)^{-2/3}(2x-1).
$$
So, indeed, the derivative is zero exactly at $x = 1/2$. Now to find the absolute minimum and maximum values, you evaluate the function at the endpoints of the interval and at the critical points, so
$$
\begin{align}
f(0) &= -1  \\
f(1) &= -1 \\
f(1/2) &= \left(\frac{1}{4} - \frac{1}{2} - 1\right)^{1/3} = \sqrt[3]{-5/4}
\end{align}
$$
The largest value is the maximum value and the smalles value is the minimum value.

If, instead, you have $f(x) = (x^2 -x \color{red}{+} 1)^{1/3}$, then the derivative is
$$
f'(x) = \frac{1}{3}(x^2 - x + 1)^{-2/3}(2x-1).
$$
Again the derivative is zero at $1/2$ and now the list of numbers is
$$
\begin{align}
f(0) &= 1 \\
f(1) &= 1 \\
f(1/2) &= \sqrt[3]{3/4}.
\end{align}
$$
Again, the absolute maximum value is the largest value in this list. The absolute minimum value if the smallest value.
One common mistake is that people forget to evaluate the original function to find the max. and min. values.
A: Since $x^2-x+1=1-x(1-x)$ and $x(1-x)\geq 0$ for $0\leq x \leq 1$. The max of your function happens when $x(1-x)$ is minimum. This minimum is clearly zero and happens at $x=0,1$. Hence $\max_{0\leq x \leq 1}(x^2-x+1)^{\frac13}=1$.
