The maximum and minimum values of {x^3} - 18{ | vedclass

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

Question

(C) Let $f(x) = {x^3} - 18{x^2} + 96x$. First,find the derivative: $f'(x) = 3{x^2} - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x

Vedclass · Accepted Answer

$160, 128$

Vedclass · Answer

(C) Let $f(x) = {x^3} - 18{x^2} + 96x$. First,find the derivative: $f'(x) = 3{x^2} - 36x + 96$. Set $f'(x) = 0$ to find critical points: $3(x^2 - 12x + 32) = 0 \Rightarrow 3(x - 4)(x - 8) = 0$. Thus,the critical points are $x = 4$ and $x = 8$. Now,find the second derivative: $f''(x) = 6x - 36$. At $x = 4$,$f''(4) = 6(4) - 36 = -12 The maximum value is $f(4) = (4)^3 - 18(4)^2 + 96(4) = 64 - 288 + 384 = 160$. At $x = 8$,$f''(8) = 6(8) - 36 = 12 > 0$,so $f(x)$ has a local minimum at $x = 8$. The minimum value is $f(8) = (8)^3 - 18(8)^2 + 96(8) = 512 - 1152 + 768 = 128$. Therefore,the maximum and minimum values are $160$ and $128$ respectively.

The maximum and minimum values of ${x^3} - 18{x^2} + 96x$ in the interval $(0, 9)$ are

Similar Questions

$A$ rectangular sheet of tin $45 \, cm$ by $24 \, cm$ is to be made into a box without a top by cutting off a square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum (in $, cm$)?

If $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has local maximum and minimum at $x = p$ and $x = q$,respectively,then $(p, q)$ equals

The number of distinct real roots of $x^4-4x^3+12x^2+x-1=0$ is

The sum of the maximum and minimum values of the function $f(x) = |5x - 7| + [x^2 + 2x]$ in the interval $[\frac{5}{4}, 2]$,where $[t]$ denotes the greatest integer function $\leq t$,is:

The maximum value of $f(x) = \frac{x}{4 + x + x^2}$ on the interval $[-1, 1]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module