The maximum value of [x(x −1) +1]^($\frac{1}{3}$) , 0 $\leq$ x $\leq$ 1 is ______.

Question

KollegeApply · Accepted Answer

The maximum value of [x(x −1) +1]^($\frac{1}{3}$) , 0 $\leq$ x $\leq$ 1 is 1. Explanation: Let, y = [x (x – 1) + 1]^($\frac{1}{3}$) Differentiating both sides with respect to x, `dy/dx = $\frac{1}{3}$ [x (x - 1) + 1]^(-$\frac{2}{3}$) d/dx [x(x - 1) + 1] = $\frac{1}{3}$ [x (x - 1) + 1]^(-$\frac{2}{3}$) $\times$ (2x - 1) = (2x - 1)/(3 [x (x - 1) + 1]^($\frac{2}{3}$))` For highest and lowest value, `dy/dx = 0 => 2x - 1 = 0 => x = $\frac{1}{2}$` For highest and lowest value, `dy/dx = 0 => 2x - 1 = 0 => x = $\frac{1}{2}$ At x= 0, f(0) = 1^($\frac{1}{3}$) = 1 At x= 1, f(1) = 1^($\frac{1}{3}$) = 1 x = $\frac{1}{2}$ at, f($\frac{1}{2}$) = [$\frac{1}{2}$ (-$\frac{1}{2}$) xx 1]^($\frac{1}{3}$) = ($\frac{3}{4}$)^($\frac{1}{3}$) dy/dx , at x = $\frac{1}{2}$` sign is changing from -ve to +ve $\therefore$ y is minimum at x = $\frac{1}{2}$. maximum value = 1

AChoose Your Answer