Questions: Given the function (f(x)=x^3-3 x^2), determine all intervals on which (f) is both decreasing and concave up.

Transcript text: Given the function $f(x)=x^{3}-3 x^{2}$, determine all intervals on which $f$ is both decreasing and concave up.

Solution

Solution Steps

Step 1: Find the First Derivative

To determine where the function $ f(x) = x^3 - 3x^2 $ is decreasing, we first need to find its first derivative:

\[ f'(x) = \frac{d}{dx}(x^3 - 3x^2) = 3x^2 - 6x \]

Step 2: Find the Critical Points

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

\[ 3x^2 - 6x = 0 \]

Factor the equation:

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

This gives the critical points $ x = 0 $ and $ x = 2 $.

Step 3: Determine Intervals of Decrease

To find where the function is decreasing, we need to test the sign of $ f'(x) $ in the intervals determined by the critical points: $ (-\infty, 0) $, $ (0, 2) $, and $ (2, \infty) $.

For $ x \in (-\infty, 0) $, choose $ x = -1 $: \[ f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 > 0 \] (not decreasing)
For $ x \in (0, 2) $, choose $ x = 1 $: \[ f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 < 0 \] (decreasing)
For $ x \in (2, \infty) $, choose $ x = 3 $: \[ f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 > 0 \] (not decreasing)

Thus, $ f(x) $ is decreasing on the interval $ (0, 2) $.

Step 4: Find the Second Derivative

To determine where the function is concave up, we find the second derivative:

\[ f''(x) = \frac{d}{dx}(3x^2 - 6x) = 6x - 6 \]

Step 5: Determine Intervals of Concavity

Set the second derivative equal to zero to find inflection points:

\[ 6x - 6 = 0 \implies x = 1 \]

Test the sign of $ f''(x) $ in the intervals $ (-\infty, 1) $ and $ (1, \infty) $:

For $ x \in (-\infty, 1) $, choose $ x = 0 $: \[ f''(0) = 6(0) - 6 = -6 < 0 \] (concave down)
For $ x \in (1, \infty) $, choose $ x = 2 $: \[ f''(2) = 6(2) - 6 = 12 - 6 = 6 > 0 \] (concave up)

Thus, $ f(x) $ is concave up on the interval $ (1, \infty) $.

Step 6: Determine the Interval Where $ f $ is Both Decreasing and Concave Up

The function $ f(x) $ is both decreasing and concave up on the interval where these two conditions overlap. From the previous steps, the overlapping interval is:

\[ (1, 2) \]