The minimum value of the function y = 2x^3 - | vedclass

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

Question

(A) Given,$f(x) = 2x^3 - 21x^2 + 36x - 20$. First,find the derivative $f'(x)$: $f'(x) = 6x^2 - 42x + 36$. To find the critical points,set $f'(x) = 0$:

Vedclass · Accepted Answer

$-128$

Vedclass · Answer

(A) Given,$f(x) = 2x^3 - 21x^2 + 36x - 20$. First,find the derivative $f'(x)$: $f'(x) = 6x^2 - 42x + 36$. To find the critical points,set $f'(x) = 0$: $6(x^2 - 7x + 6) = 0$ $6(x - 1)(x - 6) = 0$ So,$x = 1$ and $x = 6$. Now,find the second derivative $f''(x)$: $f''(x) = 12x - 42$. Check the nature of the critical points: For $x = 1$,$f''(1) = 12(1) - 42 = -30 For $x = 6$,$f''(6) = 12(6) - 42 = 72 - 42 = 30 > 0$ (Local minimum). Therefore,the minimum value occurs at $x = 6$: $f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20$ $f(6) = 2(216) - 21(36) + 216 - 20$ $f(6) = 432 - 756 + 216 - 20$ $f(6) = -128$. Thus,the minimum value is $-128$.

The minimum value of the function $y = 2x^3 - 21x^2 + 36x - 20$ is

Similar Questions

If $(2, a)$ and $(b, 19)$ are two stationary points of the curve $y=2x^3-15x^2+36x+c$,then $a+b+c=$

The coordinates of the points on the curve $4y = x^2$ that are nearest to the point $(0, 5)$ are ...

Find the local maximum and local minimum values for the function $h(x) = \sin x + \cos x$ in the interval $0 < x < \frac{\pi}{2}$.

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = x^{3}, x \in [-2, 2]$.

The curve $f(x) = e^x \sin x$ is defined in the interval $[0, 2 \pi]$. The value of $x$ for which the slope of the tangent drawn to the curve at $x$ is maximum,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module