Find the maximum value of f(x) = x^3 - 12x^2 | vedclass

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(D) Let $f(x) = x^3 - 12x^2 + 45x$.First,find the derivative $f'(x) = 3x^2 - 24x + 45 = 3(x^2 - 8x + 15) = 3(x - 3)(x - 5)$.To find the critical point

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(D) Let $f(x) = x^3 - 12x^2 + 45x$.First,find the derivative $f'(x) = 3x^2 - 24x + 45 = 3(x^2 - 8x + 15) = 3(x - 3)(x - 5)$.To find the critical points,set $f'(x) = 0$,which gives $x = 3$ and $x = 5$.Now,evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[0, 7]$:$f(0) = (0)^3 - 12(0)^2 + 45(0) = 0$$f(3) = (3)^3 - 12(3)^2 + 45(3) = 27 - 108 + 135 = 54$$f(5) = (5)^3 - 12(5)^2 + 45(5) = 125 - 300 + 225 = 50$$f(7) = (7)^3 - 12(7)^2 + 45(7) = 343 - 588 + 315 = 70$Comparing these values,the maximum value of $f(x)$ on the interval $[0, 7]$ is $70$.

Find the maximum value of $f(x) = x^3 - 12x^2 + 45x$ in the interval $[0, 7]$.

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