Determine the absolute extrema of the function $f(x) = x^3 - 12x + 1$ on the interval $[-3, 3]$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To find the critical points of the function $f(x) = x^3 - 12x + 1$, we first need to find its derivative. ‖ ‖ step 2 ⋮ The derivative of $f(x)$ is $f'(x) = 3x^2 - 12$. ‖ ‖ step 3 ⋮ Set the derivative equal to zero to find the critical points: $3x^2 - 12 = 0$. ‖ ‖ step 4 ⋮ Solve for $x$: $3(x^2 - 4) = 0 \implies x^2 - 4 = 0 \implies (x - 2)(x + 2) = 0 \implies x = \pm 2$. ‖ ‖ step 5 ⋮ The critical points are $x = 2$ and $x = -2$. ‖ ‖ step 6 ⋮ To find the absolute maximum and minimum on the interval $[-3, 3]$, evaluate $f(x)$ at the critical points and the endpoints of the interval. ‖ ‖ step 7 ⋮ Evaluate $f(x)$ at $x = -3$: $f(-3) = (-3)^3 - 12(-3) + 1 = -27 + 36 + 1 = 10$. ‖ ‖ step 8 ⋮ Evaluate $f(x)$ at $x = 3$: $f(3) = 3^3 - 12(3) + 1 = 27 - 36 + 1 = -8$. ‖ ‖ step 9 ⋮ Evaluate $f(x)$ at $x = -2$: $f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17$. ‖ ‖ step 10 ⋮ Evaluate $f(x)$ at $x = 2$: $f(2) = 2^3 - 12(2) + 1 = 8 - 24 + 1 = -15$. ‖ ‖ step 11 ⋮ Compare the values: $f(-3) = 10$, $f(3) = -8$, $f(-2) = 17$, $f(2) = -15$. ‖ ‖ step 12 ⋮ The absolute maximum value is $17$ at $x = -2$, and the absolute minimum value is $-15$ at $x = 2$. ‖ ∻Answer∻ ⚹ The absolute maximum value of $f(x)$ on the interval $[-3, 3]$ is $17$ at $x = -2$, and the absolute minimum value is $-15$ at $x = 2$. ⚹ ∻Key Concept∻ ⚹Critical points and evaluating endpoints⚹ ∻Explanation∻ ⚹To find the absolute extrema of a function on a closed interval, evaluate the function at its critical points and at the endpoints of the interval. Compare these values to determine the maximum and minimum.⚹◊What is the process for finding the absolute extrema of a function on a given interval in calculus?⍭ Generate me a similar question◊

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