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

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(B) Let $f(x) = x^{3} - 12x^{2} + 36x + 17$. Find the derivative: $f'(x) = 3x^{2} - 24x + 36$. Set $f'(x) = 0$ to find critical points: $3(x^{2} - 8x

Vedclass · Accepted Answer

$177$

Vedclass · Answer

(B) Let $f(x) = x^{3} - 12x^{2} + 36x + 17$. Find the derivative: $f'(x) = 3x^{2} - 24x + 36$. Set $f'(x) = 0$ to find critical points: $3(x^{2} - 8x + 12) = 0 \Rightarrow 3(x - 2)(x - 6) = 0$. The critical points are $x = 2$ and $x = 6$. Evaluate the function at the critical points and the interval endpoints $[1, 10]$: $f(1) = (1)^{3} - 12(1)^{2} + 36(1) + 17 = 1 - 12 + 36 + 17 = 42$. $f(2) = (2)^{3} - 12(2)^{2} + 36(2) + 17 = 8 - 48 + 72 + 17 = 49$. $f(6) = (6)^{3} - 12(6)^{2} + 36(6) + 17 = 216 - 432 + 216 + 17 = 17$. $f(10) = (10)^{3} - 12(10)^{2} + 36(10) + 17 = 1000 - 1200 + 360 + 17 = 177$. Comparing these values,the maximum value is $177$.

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

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