The maximum value of f(x) = (7-x)^4 (2+x)^5 i | vedclass

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

Question

(C) To find the maximum value,we first find the derivative $f'(x)$ and set it to $0$.$f(x) = (7-x)^4 (2+x)^5$Using the product rule and chain rule:$f'

Vedclass · Accepted Answer

$4^4 5^5$

Vedclass · Answer

(C) To find the maximum value,we first find the derivative $f'(x)$ and set it to $0$.$f(x) = (7-x)^4 (2+x)^5$Using the product rule and chain rule:$f'(x) = 4(7-x)^3(-1)(2+x)^5 + 5(7-x)^4(2+x)^4$$f'(x) = (7-x)^3(2+x)^4 [-4(2+x) + 5(7-x)]$$f'(x) = (7-x)^3(2+x)^4 [-8 - 4x + 35 - 5x]$$f'(x) = (7-x)^3(2+x)^4 [27 - 9x]$Setting $f'(x) = 0$,we get critical points $x = 7, x = -2, x = 3$.For $x \in (-2, 7)$,the function is positive. At $x = 3$,the derivative changes from positive to negative,indicating a local maximum.The maximum value is $f(3) = (7-3)^4 (2+3)^5 = 4^4 	imes 5^5$.

The maximum value of $f(x) = (7-x)^4 (2+x)^5$ is

Similar Questions

The maximum value of $\left(\frac{1}{x}\right)^x$ is

The sixth term of an $A.P.$ is equal to $2$. The value of the common difference $x$ of the $A.P.$ that makes the product $a_1 a_4 a_5$ least is given by:

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

Maximum value of the function $f(x) = [x(x-1) + 1]^{\frac{1}{3}}$ for $0 \leq x \leq 1$ is . . . . . . .

Show that the function given by $f(x) = \frac{\log x}{x}$ has a maximum at $x = e$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module