What is the minimum value of y = 2x^3 - 21x^2 | vedclass

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(A) Given function: $f(x) = 2x^3 - 21x^2 + 36x - 20$. Find the first derivative: $f'(x) = 6x^2 - 42x + 36$. Set $f'(x) = 0$ to find critical points: $

Vedclass · Accepted Answer

$-128$

Vedclass · Answer

(A) Given function: $f(x) = 2x^3 - 21x^2 + 36x - 20$. Find the first derivative: $f'(x) = 6x^2 - 42x + 36$. Set $f'(x) = 0$ to find critical points: $6(x^2 - 7x + 6) = 0 \Rightarrow 6(x - 1)(x - 6) = 0$. Thus,the critical points are $x = 1$ and $x = 6$. Find the second derivative: $f''(x) = 12x - 42$. Evaluate at 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). Calculate the minimum value at $x = 6$: $f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20$ $f(6) = 2(216) - 21(36) + 216 - 20$ $f(6) = 432 - 756 + 216 - 20 = -128$. Therefore,the minimum value is $-128$.

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

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