Question

The maximum value of [x(x-1)+1]^1/3,0≤ x≤ 1 is

Answer 1

We have to find the maximum value of $[x(x-1)+1]^{1/3},0\leq x\leq 1$.

Now consider $x(x-1)+1$.

$x(x-1)+1=x^2-x+1\\ x(x-1)+1=(x-\frac{1}{2})^2-\frac{1}{4}+1 \\ x(x-1)+1=(x-\frac{1}{2})^2+\frac{3}{4}$

The maximum value of $x(x-1)+1=(x-\frac{1}{2})^2+\frac{3}{4}$ occurs in the interval when $x=0,1$.

The maximum value of $(x-\frac{1}{2})^2+\frac{3}{4}$ is $\frac{1}{4}+\frac{3}{4}=1$. Hence the maximum value of $[x(x-1)+1]^{1/3},0\leq x\leq 1$ is $1$.