On the interval [0, 1],the function f(x) = x^ | vedclass

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

Question

(B) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum value on $[0, 1]$,we find the derivative $f'(x)$ and set it to $0$.Using the product rule: $f'(x

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(B) Let $f(x) = x^{25}(1-x)^{75}$.To find the maximum value on $[0, 1]$,we find the derivative $f'(x)$ and set it to $0$.Using the product rule: $f'(x) = 25x^{24}(1-x)^{75} + x^{25} \cdot 75(1-x)^{74}(-1)$.$f'(x) = 25x^{24}(1-x)^{74} [ (1-x) - 3x ]$.$f'(x) = 25x^{24}(1-x)^{74} (1 - 4x)$.Setting $f'(x) = 0$,we get critical points at $x = 0$,$x = 1$,and $x = 1/4$.Evaluating $f(x)$ at these points:$f(0) = 0^{25}(1-0)^{75} = 0$.$f(1) = 1^{25}(1-1)^{75} = 0$.$f(1/4) = (1/4)^{25}(3/4)^{75} > 0$.Since $f(x) \geq 0$ on $[0, 1]$ and the values at the endpoints are $0$,the maximum value occurs at $x = 1/4$.

On the interval $[0, 1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

Similar Questions

Which of the following pairs of straight lines intersect at right angles?

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is

The radius of the sphere $x^2+y^2+z^2=12x+4y+3z$ is

Which of the following statements is not correct regarding calcination?

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module