Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Monotonicity Theorem to find where the given function is increasing and where it is decreasing. $$ f(t)=t^{3}+3 t^{2}-12 $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -2)\) and \((0, \infty)\), and decreasing on \((-2, 0)\).

Step 1: Identify the Function

The given function is \( f(t) = t^3 + 3t^2 - 12 \). Our goal is to determine the intervals where this function is increasing or decreasing.

Step 2: Find the Derivative

Calculate the derivative of the function. Since \( f(t) = t^3 + 3t^2 - 12 \), the derivative is calculated as follows: \[ f'(t) = \frac{d}{dt}(t^3 + 3t^2 - 12) = 3t^2 + 6t \].

Step 3: Set the Derivative to Zero

To find critical points where the function might change from increasing to decreasing or vice versa, set the derivative equal to zero: \[ 3t^2 + 6t = 0 \].

Step 4: Solve for Critical Points

Solve the equation \( 3t^2 + 6t = 0 \) to find the values of \( t \). Factor the equation: \[ 3t(t + 2) = 0 \].So, \( t = 0 \) or \( t = -2 \). These are the critical points.

Step 5: Test Intervals

With critical points \( t = -2 \) and \( t = 0 \), we divide the number line into intervals: 1. \( (-\infty, -2) \)2. \( (-2, 0) \)3. \( (0, \infty) \)Test a point in each interval in \( f'(t) = 3t^2 + 6t \) to determine if \( f(t) \) is increasing or decreasing.

Step 6: Analyze Interval \((-\infty, -2)\)

Choose \( t = -3 \), a test point in the interval \((-\infty, -2)\): \[ f'(-3) = 3(-3)^2 + 6(-3) = 27 - 18 = 9 > 0 \] Thus, \( f(t) \) is increasing on \((-\infty, -2)\).

Step 7: Analyze Interval \((-2, 0)\)

Choose \( t = -1 \), a test point in the interval \((-2, 0)\): \[ f'(-1) = 3(-1)^2 + 6(-1) = 3 - 6 = -3 < 0 \] Thus, \( f(t) \) is decreasing on \((-2, 0)\).

Step 8: Analyze Interval \((0, \infty)\)

Choose \( t = 1 \), a test point in the interval \((0, \infty)\):\[ f'(1) = 3(1)^2 + 6(1) = 3 + 6 = 9 > 0 \]Thus, \( f(t) \) is increasing on \((0, \infty)\).

Key Concepts

Critical Points

Critical points play a key role in understanding a function's behavior. These are the points where the derivative of the function equals zero or does not exist. In our function, critical points are found by setting the derivative equal to zero and solving for the variable.
To find critical points for the given function, we first calculated the derivative:

• Function: \( f(t) = t^3 + 3t^2 - 12 \)
• Derivative: \( f'(t) = 3t^2 + 6t \)

By setting the derivative equal to zero, \( 3t^2 + 6t = 0 \), and factoring, we find:

• \( 3t(t + 2) = 0 \)
• So, \( t = 0 \) or \( t = -2 \)

These values, \( t = 0 \) and \( t = -2 \), are our critical points, indicating potential locations for changes in increasing or decreasing behavior.

Intervals of Increase and Decrease

Determining where a function is increasing or decreasing involves testing intervals defined by the critical points. Here's how you do it:
First, identify the intervals using the critical points. For \( f(t) = t^3 + 3t^2 - 12 \), we have critical points at \( t = -2 \) and \( t = 0 \). These points divide the number line into three intervals:

• \((-\infty, -2)\)
• \((-2, 0)\)
• \((0, \infty)\)

To determine the behavior of the function in each interval:

• Choose a test point within each interval.
• Evaluate the derivative \( f'(t) \) at each test point.
• If \( f'(t) > 0 \), the function is increasing.
• If \( f'(t) < 0 \), the function is decreasing.

Applying this to our function:

• Interval \((-\infty, -2)\), choose \( t = -3 \); \( f'(-3) = 9 \), function increases.
• Interval \((-2, 0)\), choose \( t = -1 \); \( f'(-1) = -3 \), function decreases.
• Interval \((0, \infty)\), choose \( t = 1 \); \( f'(1) = 9 \), function increases.

Derivative Calculation

The derivative of a function is a powerful tool for analyzing function behavior. It gives us the rate at which the function value is changing with respect to changes in the variable.
For the function \( f(t) = t^3 + 3t^2 - 12 \), we can calculate the derivative step-by-step:

• For the term \( t^3 \), differentiate to get \( 3t^2 \).
• For the term \( 3t^2 \), differentiate to get \( 6t \).
• The constant \(-12\) has a derivative of \(0\).
• Combine these to find \( f'(t) = 3t^2 + 6t \).

This derivative is essential in identifying critical points and analyzing intervals of increase and decrease. Hence, grasping the concept of differentiation is essential for function analysis.

Function Analysis

Function analysis involves a deep understanding of how a function behaves across its domain. Using tools such as derivatives, critical points, and intervals, we gain insights into the function's increasing or decreasing tendencies.
To analyze the function \( f(t) = t^3 + 3t^2 - 12 \), follow these steps:

• First, calculate the derivative, \( f'(t) = 3t^2 + 6t \), to understand the rate of change.
• Identify critical points by setting the derivative to zero, solving \( 3t^2 + 6t = 0 \), giving \( t = 0 \) and \( t = -2 \).
• Using these critical points, divide the number line into intervals and test to see if the function increases or decreases.
  • \((-\infty, -2)\), the function increases.
  • \((-2, 0)\), the function decreases.
  • \((0, \infty)\), the function increases.

This structured approach provides a complete picture of where the function rises and falls, helping predict behavior across different domains.