The maximum value of the function f(x) = x^3 | vedclass

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(B) Let $f(x) = x^3 - 12x^2 + 36x + 17$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = 3x^2 - 24x + 36$.Setting $f'(x) = 0$:

Vedclass · Accepted Answer

$177$

Vedclass · Answer

(B) Let $f(x) = x^3 - 12x^2 + 36x + 17$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = 3x^2 - 24x + 36$.Setting $f'(x) = 0$:$3(x^2 - 8x + 12) = 0$$3(x - 2)(x - 6) = 0$So,the critical points are $x = 2$ and $x = 6$.Now,we evaluate the function $f(x)$ at the critical points and the endpoints of the interval $[1, 10]$:$f(1) = (1)^3 - 12(1)^2 + 36(1) + 17 = 1 - 12 + 36 + 17 = 42$.$f(2) = (2)^3 - 12(2)^2 + 36(2) + 17 = 8 - 48 + 72 + 17 = 49$.$f(6) = (6)^3 - 12(6)^2 + 36(6) + 17 = 216 - 432 + 216 + 17 = 17$.$f(10) = (10)^3 - 12(10)^2 + 36(10) + 17 = 1000 - 1200 + 360 + 17 = 177$.Comparing these values,the maximum value is $177$ at $x = 10$.

The maximum value of the function $f(x) = x^3 - 12x^2 + 36x + 17$ in the interval $[1, 10]$ is:

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