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

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the absolute extrema of the function $f(x) = x^3 - 12x + 1$ on the interval $[-3, 3]$, we first need to find the critical points by taking the derivative of $f(x)$ and setting it to zero. ‖ ‖ step 2 ⋮ The derivative of $f(x)$ is $f'(x) = 3x^2 - 12$. Setting this equal to zero, we get $3x^2 - 12 = 0$. ‖ ‖ step 3 ⋮ Solving for $x$, we get $x^2 = 4$, so $x = \pm 2$. Thus, the critical points are $x = 2$ and $x = -2$. ‖ ‖ step 4 ⋮ Next, we evaluate the function $f(x)$ at the critical points and at the endpoints of the interval $[-3, 3]$. ‖ ‖ step 5 ⋮ Evaluating $f(x)$ at $x = -3$: $f(-3) = (-3)^3 - 12(-3) + 1 = -27 + 36 + 1 = 10$. ‖ ‖ step 6 ⋮ Evaluating $f(x)$ at $x = 3$: $f(3) = 3^3 - 12(3) + 1 = 27 - 36 + 1 = -8$. ‖ ‖ step 7 ⋮ Evaluating $f(x)$ at $x = -2$: $f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17$. ‖ ‖ step 8 ⋮ Evaluating $f(x)$ at $x = 2$: $f(2) = 2^3 - 12(2) + 1 = 8 - 24 + 1 = -15$. ‖ ‖ step 9 ⋮ Comparing these values, we find that the maximum value of $f(x)$ on the interval $[-3, 3]$ is $17$ at $x = -2$, and the 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∻ ⚹Absolute Extrema⚹ ∻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 values.⚹◊What method can be used to find the absolute extrema of a function within⍭ Generate me a similar question◊

Sia