The minimum value of the function f(x) = 3x^4 | vedclass

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

Question

(B) Given the function $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$.First,find the derivative $f'(x)$:$f'(x) = 12x^3 - 24x^2 + 24x - 48$$f'(x) = 12(x^3 - 2

Vedclass · Accepted Answer

$-39$

Vedclass · Answer

(B) Given the function $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$.First,find the derivative $f'(x)$:$f'(x) = 12x^3 - 24x^2 + 24x - 48$$f'(x) = 12(x^3 - 2x^2 + 2x - 4)$$f'(x) = 12[x^2(x - 2) + 2(x - 2)] = 12(x - 2)(x^2 + 2)$.To find critical points,set $f'(x) = 0$:Since $x^2 + 2 = 0$ has no real roots,the only critical point is $x = 2$.Now,evaluate $f(x)$ at the critical point $x = 2$ and the endpoints of the interval $[0, 3]$:$f(0) = 3(0)^4 - 8(0)^3 + 12(0)^2 - 48(0) + 25 = 25$$f(2) = 3(2)^4 - 8(2)^3 + 12(2)^2 - 48(2) + 25 = 3(16) - 8(8) + 12(4) - 96 + 25 = 48 - 64 + 48 - 96 + 25 = -39$$f(3) = 3(3)^4 - 8(3)^3 + 12(3)^2 - 48(3) + 25 = 3(81) - 8(27) + 12(9) - 144 + 25 = 243 - 216 + 108 - 144 + 25 = 16$.The minimum value is $\min\{25, -39, 16\} = -39$.

The minimum value of the function $f(x) = 3x^4 - 8x^3 + 12x^2 - 48x + 25$ on the interval $[0, 3]$ is equal to:

Similar Questions

Find two numbers whose sum is $24$ and whose product is as large as possible.

The area bounded by the tangents of the curve given by $y = \sin \theta \cos^2 \theta$ and $x = \sin^2 \theta \cos \theta$,which are parallel to the coordinate axes (excluding the axes themselves),is:

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

The minimum value of the function $f(x) = x^{10} + x^2 + \frac{1}{x^{12}} + \frac{1}{1 + \sec^{-1} x}$ is

Point $A$ lies on the curve $y = e^{-x^2}$ and has the coordinates $(x, e^{-x^2})$ where $x > 0$. Point $B$ has the coordinates $(x, 0)$. If $O$ is the origin,then the maximum area of the triangle $AOB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module