Questions: Consider the following. f(x) = 1/4 x^4 - 1/3 x^3 - x^2 Find the critical numbers. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) x = 0.1, -2 Find the open intervals on which the function is increasing or decreasing. Use a graphing utility to verify your results. (Enter your answers using interval notation, If an answer does not exist, enter DNE.) Increasing (-∞, -2), (1, ∞) Decreasing (-2, 1)

Consider the following.
f(x) = 1/4 x^4 - 1/3 x^3 - x^2

Find the critical numbers. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
x = 0.1, -2

Find the open intervals on which the function is increasing or decreasing. Use a graphing utility to verify your results. (Enter your answers using interval notation, If an answer does not exist, enter DNE.)
Increasing (-∞, -2), (1, ∞)
Decreasing (-2, 1)

Transcript text: Consider the following. \[ f(x)=\frac{1}{4} x^{4}-\frac{1}{3} x^{3}-x^{2} \] Find the critical numbers. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) \[ x=0.1,-2 \] Find the open intervals on which the function is increasing or decreasing. Use a graphing utility to verify your results. (Enter your answers using interval notation, If an answer does not exist, enter DNE.) Increasing $\quad(-\infty,-2),(1, \infty)$ decreasing \[ (-2,1) \]

Solution

Solution Steps

To find the critical numbers of the function, we first need to find the derivative of the function and set it equal to zero. The solutions to this equation are the critical numbers. For determining the intervals where the function is increasing or decreasing, we analyze the sign of the derivative on these intervals. If the derivative is positive, the function is increasing; if negative, the function is decreasing.

Step 1: Find the Derivative of the Function

To find the critical numbers, we first need to find the derivative of the function $ f(x) = \frac{1}{4}x^4 - \frac{1}{3}x^3 - x^2 $.

The derivative $ f'(x) $ is calculated as follows:

\[ f'(x) = \frac{d}{dx}\left(\frac{1}{4}x^4\right) - \frac{d}{dx}\left(\frac{1}{3}x^3\right) - \frac{d}{dx}(x^2) \]

\[ f'(x) = \frac{1}{4} \cdot 4x^3 - \frac{1}{3} \cdot 3x^2 - 2x \]

\[ f'(x) = x^3 - x^2 - 2x \]

Step 2: Find the Critical Numbers

Critical numbers occur where the derivative is zero or undefined. Since $ f'(x) = x^3 - x^2 - 2x $ is a polynomial, it is defined everywhere. We set the derivative equal to zero to find the critical numbers:

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

Factor the equation:

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

Solve for $ x $:

$ x = 0 $
Solve $ x^2 - x - 2 = 0 $ using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where $ a = 1 $, $ b = -1 $, $ c = -2 $.

\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \]

\[ x = \frac{1 \pm \sqrt{1 + 8}}{2} \]

\[ x = \frac{1 \pm \sqrt{9}}{2} \]

\[ x = \frac{1 \pm 3}{2} \]

\[ x = 2 \quad \text{or} \quad x = -1 \]

Thus, the critical numbers are $ x = 0, -1, 2 $.

Step 3: Determine Intervals of Increase and Decrease

To determine where the function is increasing or decreasing, we analyze the sign of $ f'(x) = x^3 - x^2 - 2x $ on the intervals determined by the critical numbers $ x = -1, 0, 2 $.

Interval $(- \infty, -1)$: Choose $ x = -2 $. \[ f'(-2) = (-2)^3 - (-2)^2 - 2(-2) = -8 - 4 + 4 = -8 \quad (\text{negative}) \] The function is decreasing on $(- \infty, -1)$.
Interval $(-1, 0)$: Choose $ x = -0.5 $. \[ f'(-0.5) = (-0.5)^3 - (-0.5)^2 - 2(-0.5) = -0.125 - 0.25 + 1 = 0.625 \quad (\text{positive}) \] The function is increasing on $(-1, 0)$.
Interval $(0, 2)$: Choose $ x = 1 $. \[ f'(1) = 1^3 - 1^2 - 2(1) = 1 - 1 - 2 = -2 \quad (\text{negative}) \] The function is decreasing on $(0, 2)$.
Interval $(2, \infty)$: Choose $ x = 3 $. \[ f'(3) = 3^3 - 3^2 - 2(3) = 27 - 9 - 6 = 12 \quad (\text{positive}) \] The function is increasing on $(2, \infty)$.