Questions: Determine the intervals over which f(x)=x^4-4x^3+12 is increasing or decreasing.

Transcript text: Determine the intervals over which $f(x)=x^{4}-4 x^{3}+12$ is increasing or decreasing.

Solution

Step 1: Find the First Derivative

To determine where the function $ f(x) = x^4 - 4x^3 + 12 $ is increasing or decreasing, we first need to find its first derivative, $ f'(x) $.

\[ f'(x) = \frac{d}{dx}(x^4 - 4x^3 + 12) = 4x^3 - 12x^2 \]

Step 2: Find Critical Points

Set the first derivative equal to zero to find the critical points:

\[ 4x^3 - 12x^2 = 0 \]

Factor out the greatest common factor:

\[ 4x^2(x - 3) = 0 \]

This gives us the critical points:

\[ x = 0 \quad \text{and} \quad x = 3 \]

Step 3: Determine Intervals of Increase and Decrease

We will test the intervals determined by the critical points: $(-\infty, 0)$, $(0, 3)$, and $(3, \infty)$.

Interval $(-\infty, 0)$: Choose $ x = -1 $. \[ f'(-1) = 4(-1)^2(-1 - 3) = 4 \cdot 1 \cdot (-4) = -16 \] Since $ f'(-1) < 0 $, $ f(x) $ is decreasing on $(-\infty, 0)$.
Interval $(0, 3)$: Choose $ x = 1 $. \[ f'(1) = 4(1)^2(1 - 3) = 4 \cdot 1 \cdot (-2) = -8 \] Since $ f'(1) < 0 $, $ f(x) $ is decreasing on $(0, 3)$.
Interval $(3, \infty)$: Choose $ x = 4 $. \[ f'(4) = 4(4)^2(4 - 3) = 4 \cdot 16 \cdot 1 = 64 \] Since $ f'(4) > 0 $, $ f(x) $ is increasing on $(3, \infty)$.