Question

find the exact location of the relative and absolute extrema of the function $g(x) = -3x^3 + 9x + 6$ on the interval $-3, 3$. (enter dne for \does not exist\ if there are none.)
$g(x)$ has an absolute minimum at $(x, y) = (\underline{\quad}, \underline{\quad})$.
$g(x)$ has an absolute maximum at $(x, y) = (\underline{\quad}, \underline{\quad})$.
$g(x)$ has a relative minimum at $(x, y) = (\underline{\quad}, \underline{\quad})$.
$g(x)$ has a relative maximum at $(x, y) = (\underline{\quad}, \underline{\quad})$.

Explanation:

Step1: Find first derivative

$g'(x) = \frac{d}{dx}(-3x^3 + 9x + 6) = -9x^2 + 9$

Step2: Find critical points

Set $g'(x)=0$:
$-9x^2 + 9 = 0$
$9x^2 = 9$
$x^2 = 1$
$x = 1, x = -1$

Step3: Second derivative test

$g''(x) = \frac{d}{dx}(-9x^2 + 9) = -18x$

• For $x=1$: $g''(1) = -18(1) = -18 < 0$, so relative maximum at $x=1$.
• For $x=-1$: $g''(-1) = -18(-1) = 18 > 0$, so relative minimum at $x=-1$.

Step4: Compute $g(x)$ at critical points and endpoints

• At $x=-3$: $g(-3) = -3(-3)^3 + 9(-3) + 6 = 81 - 27 + 6 = 60$
• At $x=-1$: $g(-1) = -3(-1)^3 + 9(-1) + 6 = 3 - 9 + 6 = 0$
• At $x=1$: $g(1) = -3(1)^3 + 9(1) + 6 = -3 + 9 + 6 = 12$
• At $x=3$: $g(3) = -3(3)^3 + 9(3) + 6 = -81 + 27 + 6 = -48$

Step5: Identify absolute extrema

• Absolute minimum is the smallest value: $g(3)=-48$
• Absolute maximum is the largest value: $g(-3)=60$

Answer:

$g(x)$ has an absolute minimum at $(x,y) = (3, -48)$.
$g(x)$ has an absolute maximum at $(x,y) = (-3, 60)$.
$g(x)$ has a relative minimum at $(x,y) = (-1, 0)$.
$g(x)$ has a relative maximum at $(x,y) = (1, 12)$.