Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-10, 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 \((-\infty, -2) \cup (0, \infty)\) and decreasing on \((-2, 0)\).

Step 1: Find the First Derivative

To apply the Monotonicity Theorem, we first need to find the derivative of the given function. The function is \( f(t) = t^3 + 3t^2 - 12 \). Using the power rule, the derivative \( f'(t) \) can be calculated as follows:\[f'(t) = \frac{d}{dt}(t^3) + \frac{d}{dt}(3t^2) + \frac{d}{dt}(-12)\]This simplifies to:\[f'(t) = 3t^2 + 6t\]

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. The derivative found is \( 3t^2 + 6t \). Set this equal to zero to find the critical points:\[3t^2 + 6t = 0\]Factor the equation:\[3t(t + 2) = 0\]This gives the critical points \( t = 0 \) and \( t = -2 \).

Step 3: Test Intervals Around Critical Points

We need to determine whether the function is increasing or decreasing around the critical points. Consider the intervals around the critical points: \((-\infty, -2)\), \((-2, 0)\), and \((0, \infty)\).1. For \( t \in (-\infty, -2) \): Choose \( t = -3 \), then \( f'(-3) = 3((-3)^2) + 6(-3) = 27 - 18 = 9 > 0 \). Thus, \( f(t) \) is increasing.2. For \( t \in (-2, 0) \): Choose \( t = -1 \), then \( f'(-1) = 3((-1)^2) + 6(-1) = 3 - 6 = -3 < 0 \). Thus, \( f(t) \) is decreasing.3. For \( t \in (0, \infty) \): Choose \( t = 1 \), then \( f'(1) = 3(1^2) + 6(1) = 3 + 6 = 9 > 0 \). Thus, \( f(t) \) is increasing.

Step 4: Conclude About the Function's Monotonicity

Based on our interval tests:- The function \( f(t) \) is increasing on the intervals \((-\infty, -2)\) and \((0, \infty)\).- The function is decreasing on the interval \((-2, 0)\).

Key Concepts

First Derivative

The first derivative of a function is like a window into the behavior of that function. By calculating the first derivative, noted as \( f'(t) \), we are determining the slope of the tangent line at any point \( t \). This slope tells us how the function is changing at that point.

In the context of our exercise, we started with \( f(t) = t^3 + 3t^2 - 12 \). Using the power rule, which states that the derivative of \( t^n \) is \( nt^{n-1} \), we derive \( f'(t) = 3t^2 + 6t \).

Understanding this derivative is crucial to the rest of the process because it paves the way to finding where the function changes its nature from increasing to decreasing, or vice versa.

Critical Points

Critical points are key in understanding a function's behavior. These are the values of \( t \) where the derivative \( f'(t) \) equals zero or does not exist. Why are these points so important? Because they are the potential areas where the function might change direction.

To find these points, we set our first derivative \( 3t^2 + 6t \) equal to zero and solve for \( t \).

• We factor the equation: \( 3t(t + 2) = 0 \).
• This gives us critical points at \( t = 0 \) and \( t = -2 \).

These critical points will be our guideposts for the next step: testing intervals around them to determine the nature of the function in those regions.

Increasing and Decreasing Functions

Once critical points are established, the next step is to identify where the function is increasing or decreasing. This characteristic of a function lets us understand its overall behavior across intervals.

A function is increasing on an interval if its derivative is positive throughout that interval, which means the function rises as \( t \) moves along the interval. Conversely, a function is decreasing if its derivative is negative, implying the function falls as \( t \) advances.

Following this, we make use of interval testing to see how \( f(t) \) behaves in different regions demarcated by our critical points.

Interval Testing

Interval testing is the method by which we probe the behavior of a function in the various segments created by critical points. For our function \( f(t) = t^3 + 3t^2 - 12 \), these segments are \((-\infty, -2)\), \((-2, 0)\), and \((0, \infty)\).

Here's how we test each interval:

• For \( t \in (-\infty, -2) \), a test point \( t = -3 \) gives \( f'(-3) = 9 \), indicating \( f(t) \) is increasing.
• For \( t \in (-2, 0) \), a test point \( t = -1 \) yields \( f'(-1) = -3 \), showing \( f(t) \) is decreasing.
• For \( t \in (0, \infty) \), a test point \( t = 1 \) results in \( f'(1) = 9 \), confirming \( f(t) \) is again increasing.

By performing these tests, we clearly identify where our function increases and decreases, enabling us to sketch an accurate picture of its behavior.