The function f(x) = 2x^3 - 21x^2 + 36x + 7 ha | vedclass

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

Question

(B) Given function is $f(x) = 2x^3 - 21x^2 + 36x + 7$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x + 7) = 6x^2 - 4

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given function is $f(x) = 2x^3 - 21x^2 + 36x + 7$.First,find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x + 7) = 6x^2 - 42x + 36$.Set $f'(x) = 0$ to find critical points:$6(x^2 - 7x + 6) = 0$$6(x - 1)(x - 6) = 0$So,$x = 1$ and $x = 6$.Now,find the second derivative $f''(x)$:$f''(x) = \frac{d}{dx}(6x^2 - 42x + 36) = 12x - 42$.Check the sign of $f''(x)$ at critical points:For $x = 1$: $f''(1) = 12(1) - 42 = -30$. Since $f''(1) For $x = 6$: $f''(6) = 12(6) - 42 = 72 - 42 = 30$. Since $f''(6) > 0$,the function has a local minimum at $x = 6$.Therefore,the function has a local maximum at $x = 1$.

The function $f(x) = 2x^3 - 21x^2 + 36x + 7$ has a local maximum at $x = $ ........

Similar Questions

$A$ wire of length $20 \text{ cm}$ is bent in the form of a sector of a circle. The maximum area that can be enclosed by the wire is (in $\text{ cm}^2$)

If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively,then $2 a+b$ is equal to

The triangle of maximum area that can be inscribed in a given circle of radius $r$ is ...... .

The difference between the greatest and least values of the function $f(x) = -x + \sin 2x$ on $[-\frac{\pi}{2}, \frac{\pi}{2}]$ is

Find the maximum and minimum values of the function $f(x)=(2x-1)^{2}+3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module