The function f(x) = 2x^3 - 15x^2 + 36x + 4 is | vedclass

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

Question

(A) Given function is $f(x) = 2x^3 - 15x^2 + 36x + 4$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 4) = 6x^2 - 3

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Given function is $f(x) = 2x^3 - 15x^2 + 36x + 4$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 4) = 6x^2 - 30x + 36$.To find the critical points,set $f'(x) = 0$:$6(x^2 - 5x + 6) = 0$$6(x - 2)(x - 3) = 0$So,the critical points are $x = 2$ and $x = 3$.Now,find the second derivative $f''(x)$ to check for maxima or minima:$f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30$.Evaluate $f''(x)$ at the critical points:For $x = 2$: $f''(2) = 12(2) - 30 = 24 - 30 = -6$.Since $f''(2) For $x = 3$: $f''(3) = 12(3) - 30 = 36 - 30 = 6$.Since $f''(3) > 0$,the function has a local minimum at $x = 3$.Therefore,the function is maximum at $x = 2$.

The function $f(x) = 2x^3 - 15x^2 + 36x + 4$ is maximum at $x=$ ......

Similar Questions

What are the minimum and maximum values of the function $f(x) = x^5 - 5x^4 + 5x^3 - 10$?

The denominator of a fraction is $16$ more than the square of the numerator. Find the least value of the fraction.

If the functions $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2$,where $a \neq 2b$,have a common extreme point,then $a + 2b + 7$ is equal to

The maximum value of $f(x) = (x + 1)^{\frac{1}{3}} - (x - 1)^{\frac{1}{3}}$ for $x \in [0, 1]$ is ....

If the function $f(x) = a \sin(x) + \frac{1}{3} \sin(3x)$ attains its maximum value at $x = \frac{\pi}{3}$,then $a$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module