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

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(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

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