Problem 6 - Calculus

Question

In Problems \(I-10\), 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: Understand the Function

We have the function \( f(t)=t^{3}+3t^{2}-12 \). To determine where the function is increasing or decreasing, we will need to find the derivative of the function.

Step 2: Differentiate the Function

Find the first derivative of the function \( f(t) \). Applying the power rule, the derivative is \( f'(t) = 3t^2 + 6t \).

Step 3: Find Critical Points

To find critical points, set the derivative equal to zero: \( 3t^2 + 6t = 0 \). Factor the equation to get \( 3t(t + 2) = 0 \). Thus, the solutions are \( t = 0 \) and \( t = -2 \). These are the critical points.

Step 4: Test Intervals Around Critical Points

Now, determine where the function is increasing or decreasing by testing intervals around the critical points. Choose test points in the intervals (e.g., \( t = -3 \), \( t = -1 \), and \( t = 1 \)) to substitute into \( f'(t) = 3t^2 + 6t \).

Step 5: Analyze the Intervals

- For \( t < -2 \), test \( t = -3 \): \( f'(-3) = 3(-3)^2 + 6(-3) = 27 - 18 = 9 > 0 \), so the function is increasing.- For \( -2 < t < 0 \), test \( t = -1 \): \( f'(-1) = 3(-1)^2 + 6(-1) = 3 - 6 = -3 < 0 \), so the function is decreasing.- For \( t > 0 \), test \( t = 1 \): \( f'(1) = 3(1)^2 + 6(1) = 3 + 6 = 9 > 0 \), so the function is increasing.

Step 6: Conclusion

Using the Monotonicity Theorem, we conclude that \( f(t) \) is increasing on \( (-\infty, -2) \cup (0, \infty) \) and decreasing on \( (-2, 0) \).

Key Concepts

Derivative

In calculus, the derivative of a function tells us how the function's output value changes as the input changes. In simpler terms, it measures the rate of change or the slope of a function at any given point. To find where a function is increasing or decreasing, finding its derivative is the first essential step.

For example, with the function \( f(t) = t^3 + 3t^2 - 12 \), we apply the power rule to find the derivative, which gives us \( f'(t) = 3t^2 + 6t \). Here, the derivative \( f'(t) \) represents the slope of the function \( f(t) \) at any point \( t \). This helps us understand how the function behaves across different intervals, which leads us to the next step: finding critical points.

Critical Points

Critical points of a function are those input values where the derivative is zero or undefined. These points are crucial because they potentially indicate a change in the function's increasing or decreasing behavior.

For the function \( f(t) = t^3 + 3t^2 - 12 \), we found the derivative to be \( f'(t) = 3t^2 + 6t \). Setting the derivative to zero helps locate these critical points:

• \( 3t^2 + 6t = 0 \)
• Factoring gives \( 3t(t + 2) = 0 \)
• Thus, the critical points are \( t = 0 \) and \( t = -2 \)

These values are where the slope of the function is zero, meaning there could be a shift from increasing to decreasing, or vice versa.

Interval Testing

After finding the critical points, the next step is interval testing. This means checking how the function behaves on the intervals defined by these critical points. It's a key method to determine where a function is increasing or decreasing.

For our function, we have critical points at \( t = 0 \) and \( t = -2 \). This divides the number line into intervals:

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

By selecting a test point from each interval and plugging it into the derivative \( f'(t) = 3t^2 + 6t \), we can determine whether the original function is increasing (positive derivative) or decreasing (negative derivative) in that interval.

Increasing and Decreasing Functions

Understanding where a function increases or decreases is vital for analyzing its behavior, and it's precisely why we use concepts like derivatives and critical points.

In the case of the function \( f(t) = t^3 + 3t^2 - 12 \):

• In the interval \( (-\infty, -2) \), evaluations show the derivative \( f'(t) > 0 \), indicating the function is increasing.
• In the interval \( (-2, 0) \), with \( f'(t) < 0 \), the function is decreasing.
• In the interval \( (0, \infty) \), \( f'(t) > 0 \), once again showing the function is increasing.

By using interval testing and derivatives, we can make well-reasoned conclusions about where our function is moving upwards or downwards, helping us understand its overall trend and shape.