Question

for the following exercise, use a graphing utility to estimate the local extrema of the function and to estimate the intervals on which the function is increasing and decreasing.
$f(x) = x^4 - 12x^3 + 2$
local extrema:
increasing on the interval:
decreasing on the interval:
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Explanation:

Step1: Find critical points via derivative

First, calculate the derivative of $f(x)$:
$f'(x) = 4x^3 - 36x^2$
Set $f'(x)=0$ to find critical points:
$4x^2(x - 9) = 0$
Solutions: $x=0$ and $x=9$

Step2: Test intervals for monotonicity

• For $x < 0$: Choose $x=-1$, $f'(-1)=4(-1)^3 -36(-1)^2=-4-36=-40<0$, function is decreasing.
• For $0 < x < 9$: Choose $x=1$, $f'(1)=4(1)^3 -36(1)^2=4-36=-32<0$, function is decreasing.
• For $x > 9$: Choose $x=10$, $f'(10)=4(10)^3 -36(10)^2=4000-3600=400>0$, function is increasing.

Step3: Identify local extrema

• At $x=0$: The function does not change direction (stays decreasing), so no extremum here.
• At $x=9$: Function changes from decreasing to increasing, so this is a local minimum. Calculate $f(9)$:

$f(9)=9^4 -12(9)^3 +2=6561 - 12(729)+2=6561-8748+2=-2185$

Answer:

Local extrema: (local minimum at $x=9$, $f(9)=-2185$)
Increasing on the interval: $(9, \infty)$
Decreasing on the interval: $(-\infty, 9)$