The maximum value of [x(x-1)+1]^{frac{1}{3}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = [x(x-1)+1]^{\frac{1}{3}} = [x^2-x+1]^{\frac{1}{3}}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = \frac{1}{

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Let $f(x) = [x(x-1)+1]^{\frac{1}{3}} = [x^2-x+1]^{\frac{1}{3}}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = \frac{1}{3}[x^2-x+1]^{-\frac{2}{3}} \cdot (2x-1) = \frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}$.Setting $f'(x) = 0$,we get $2x-1 = 0$,which implies $x = \frac{1}{2}$.Since $x = \frac{1}{2}$ lies within the interval $[0, 1]$,we evaluate $f(x)$ at the critical point and the endpoints:$f(0) = [0(0-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(1) = [1(1-1)+1]^{\frac{1}{3}} = 1^{\frac{1}{3}} = 1$.$f(\frac{1}{2}) = [\frac{1}{2}(\frac{1}{2}-1)+1]^{\frac{1}{3}} = [\frac{1}{2}(-\frac{1}{2})+1]^{\frac{1}{3}} = [-\frac{1}{4}+1]^{\frac{1}{3}} = (\frac{3}{4})^{\frac{1}{3}}$.Comparing the values $1$,$1$,and $(\frac{3}{4})^{\frac{1}{3}}$,the maximum value is $1$.Thus,the correct answer is $A$.

The maximum value of $[x(x-1)+1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is

Similar Questions

The following figure shows the graph of a differentiable function $y=f(x)$ on the interval $[a, b]$ (not containing $0$). Let $g(x)=\frac{f(x)}{x}$. Which of the following is a possible graph of $y=g(x)$?

Let $f(x) = (x - 3)^{2018}(x - 2)^{2019}, x \in R$. If $f(\alpha)$ is a relative maximum of $f$ at $x = \alpha$, then $2\alpha + 3f(\alpha) =$

$A$ window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is $10 \, m$. Find the dimensions of the window to admit maximum light through the whole opening.

For all real $x$,the minimum value of $\frac{1-x+x^{2}}{1+x+x^{2}}$ is

$A$ rectangle with its sides parallel to the $X$-axis and $Y$-axis is inscribed in the region bounded by the curves $y=x^2-4$ and $y=\frac{4-x^2}{2}$. The maximum possible area of such a rectangle is closest to the integer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module