Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). $$ g(x)=4 x^{3}-3 x^{2}-6 x+12 $$

Step-by-Step Solution

Answer

Increasing: \((-\infty, -\frac{1}{2}) \cup (1, \infty)\), Decreasing: \((-\frac{1}{2}, 1)\). Concave up: \((\frac{1}{4}, \infty)\), Concave down: \((-\infty, \frac{1}{4})\).

Step 1: Find the First Derivative

To determine where the function is increasing or decreasing, we first need to find the first derivative of the function. The given function is \( g(x)=4x^3-3x^2-6x+12 \). Using derivative rules:\[ g'(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(6x) + \frac{d}{dx}(12) \] \[ g'(x) = 12x^2 - 6x - 6 \].

Step 2: Determine Critical Points

Set the first derivative \( g'(x) = 12x^2 - 6x - 6 \) to zero to find critical points. Solve the equation: \[ 12x^2 - 6x - 6 = 0 \]. Simplify by dividing all terms by 6: \[ 2x^2 - x - 1 = 0 \]. Now, factor the quadratic: \[ (2x + 1)(x - 1) = 0 \]. The solutions are \( x = -\frac{1}{2} \) and \( x = 1 \).

Step 3: Test Intervals for Increasing/Decreasing

Choose test points in each interval divided by the critical points (\( x = -\frac{1}{2} \) and \( x = 1 \)) to determine where the function is increasing or decreasing: \((-\infty, -\frac{1}{2})\), \((-\frac{1}{2}, 1)\), and \((1, \infty)\). 1. Test \(x = -1\) for \((-\infty, -\frac{1}{2})\): \(g'(-1) = 12(-1)^2 - 6(-1) - 6 = 12 > 0\), increasing. 2. Test \(x = 0\) for \((-\frac{1}{2}, 1)\): \(g'(0) = -6 < 0\), decreasing. 3. Test \(x = 2\) for \((1, \infty)\): \(g'(2) = 12(2)^2 - 6(2) - 6 = 30 > 0\), increasing.

Step 4: Find the Second Derivative for Concavity

To find concave up and concave down intervals, determine the second derivative of \( g(x) \). Start from \( g'(x) = 12x^2 - 6x - 6 \):\[ g''(x) = \frac{d}{dx}(12x^2) - \frac{d}{dx}(6x) - \frac{d}{dx}(6) \] \[ g''(x) = 24x - 6 \].

Step 5: Determine Points of Inflection

Set the second derivative \( g''(x) = 24x - 6 \) to zero to find potential inflection points. Solve: \[ 24x - 6 = 0 \]. Thus, \( x = \frac{1}{4} \). This is the point where concavity changes.

Step 6: Test Intervals for Concavity

Test the intervals determined by \( x = \frac{1}{4} \): \((-\infty, \frac{1}{4})\) and \((\frac{1}{4}, \infty)\). 1. Test \(x = 0\) for \((-\infty, \frac{1}{4})\): \(g''(0) = -6 < 0\), concave down. 2. Test \(x = 1\) for \((\frac{1}{4}, \infty)\): \(g''(1) = 18 > 0\), concave up.

Key Concepts

First Derivative

The first derivative of a function helps us understand the rate of change or slope at any given point. By finding the first derivative, we can determine where a function is increasing or decreasing. For this particular function, the first derivative is calculated using standard differentiation rules. We start with the function:

• Given Function: \( g(x) = 4x^3 - 3x^2 - 6x + 12 \)
• Derivative: \( g'(x) = 12x^2 - 6x - 6 \) through term-by-term differentiation.

When we calculate this derivative, we effectively find a formula for the slope of the tangent to the graph at any point \( x \).

When \( g'(x) > 0 \), the function is increasing, indicating that as \( x \) increases, \( g(x) \) also increases. Conversely, when \( g'(x) < 0 \), the function is decreasing.

Critical Points

Critical points provide insights into where the function may change its increasing or decreasing behavior. They occur where the first derivative equals zero or is undefined.
Finding critical points involves solving \( g'(x) = 0 \):

• Derivative: \( 12x^2 - 6x - 6 = 0 \), simplified to \( 2x^2 - x - 1 = 0 \).
• After factoring, we find the solutions: \( x = -\frac{1}{2} \) and \( x = 1 \).

These points are potential locations where the function transitions from increasing to decreasing or vice versa. To confirm these transitions, test values in intervals around these critical points to check the sign of \( g'(x) \). For this function:

• For \((-\infty, -\frac{1}{2})\), \( g'(x) > 0 \), indicating the function is increasing.
• For \((-\frac{1}{2}, 1)\), \( g'(x) < 0 \), indicating the function is decreasing.
• For \((1, \infty)\), \( g'(x) > 0 \) again, indicating it increases after \( x = 1 \).

Second Derivative

The second derivative gives us information about the concavity of the function, telling us whether it bends upwards or downwards. This is crucial for determining points of inflection, where concavity changes.
To find the second derivative, take the derivative of \( g'(x) = 12x^2 - 6x - 6 \):

• Second Derivative: \( g''(x) = 24x - 6 \).

Points where \( g''(x) \) is zero or undefined could signify a change in concavity. Solving \( g''(x) = 0 \) gives possible inflection points.

• Here, \( 24x - 6 = 0 \) gives \( x = \frac{1}{4} \) as the point where the concavity potentially changes.

Concavity

Concavity helps us understand how the graph of the function is curving. A function is concave up where the graph bends upwards like a cup, and concave down where it bends downwards like a cap.
To determine concavity, we use the sign of the second derivative, \( g''(x) = 24x - 6 \):

• If \( g''(x) > 0 \), the function is concave up.
• If \( g''(x) < 0 \), the function is concave down.

Testing intervals divided by the inflection point \( x = \frac{1}{4} \):

• In \((-\infty, \frac{1}{4})\), testing \( x = 0 \) shows \( g''(0) = -6 \), indicating that the graph is concave down.
• In \((\frac{1}{4}, \infty)\), testing \( x = 1 \) gives \( g''(1) = 18 \), indicating concave up.

This means the function changes from concave down to concave up at \( x = \frac{1}{4} \), showcasing a smooth transition in the graphical curvature.