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

(C) Let $f(x) = x^{25}(1-x)^{75}$.To find the critical points,we differentiate $f(x)$ with respect to $x$ using the product rule:$f'(x) = 25x^{24}(1-x

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(C) Let $f(x) = x^{25}(1-x)^{75}$.To find the critical points,we differentiate $f(x)$ with respect to $x$ using the product rule:$f'(x) = 25x^{24}(1-x)^{75} + x^{25} \cdot 75(1-x)^{74} \cdot (-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 $x = \frac{1}{4}$.Since $f(0) = 0$ and $f(1) = 0$,and $f(x) > 0$ for $x \in (0,1)$,the function must attain its maximum at the critical point $x = \frac{1}{4}$.

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

Similar Questions

Let $f(x)=(x+3)^2(x-2)^3$ for $x \in [-4,4]$. If $M$ and $m$ are the maximum and minimum values of $f$ respectively in $[-4,4]$,then the value of $M-m$ is:

$A$ YouTube short video is going viral according to the function $f(t) = -2t^3 + 3t^2 + 5$. At what time $t$ (in hours) does the video get the maximum number of shares?

The function $f(x) = x e^{-x}, \forall x \in R$ attains a maximum value at $x$ equal to:

The maximum area of a rectangle with a perimeter of $176 \ cm$ is .......... $sq. \ cm$.

Let $f(x)=(x-2)^{17}(x+5)^{24}$. Then

Vedclass Test Series

Exam Paper Generator

Online Exam Module