Problem 9 - Thomas Calculus

Question

In Exercises \(9-28 :\) a. Find the intervals on which the function is increasing and decreasing. b. Then identify the function's local extreme values, if any, saying where they are taken on. c. Which, if any, of the extreme values are absolute? d. Support your findings with a graphing calculator or computer grapher. $$ g(t)=-t^{2}-3 t+3 $$

Step-by-Step Solution

Answer

The function increases on \((- fty, -\frac{3}{2})\) and decreases on \((-\frac{3}{2}, fty)\). It has a local maximum at \( t = -\frac{3}{2} \), but no absolute extremes.

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we first need to find its derivative. The function given is \( g(t) = -t^2 - 3t + 3 \). The derivative of \( g(t) \) with respect to \( t \) is \( g'(t) = -2t - 3 \).

Step 2: Critical Points

We identify critical points by setting the derivative equal to zero and solving for \( t \). Thus, \( -2t - 3 = 0 \) leads to \( -2t = 3 \), giving \( t = -\frac{3}{2} \). This is the critical point of the function.

Step 3: Test Intervals

We analyze the intervals around the critical point \( t = -\frac{3}{2} \) to determine if the function is increasing or decreasing. Choose test points such as \( t = -2 \) and \( t = 0 \). - For \( t = -2 \), \( g'(-2) = -2(-2) - 3 = 4 - 3 = 1 \) (positive), indicating the function is increasing on \((-fty, -\frac{3}{2})\).- For \( t = 0 \), \( g'(0) = -2(0) - 3 = -3 \) (negative), indicating the function is decreasing on \((-\frac{3}{2}, fty)\).

Step 4: Local Extreme Values

Using the First Derivative Test, a change from increasing to decreasing at \( t = -\frac{3}{2} \) indicates a local maximum. To find the value, substitute \( t = -\frac{3}{2} \) into \( g(t) \).\[ g(-\frac{3}{2}) = -(-\frac{3}{2})^2 - 3(-\frac{3}{2}) + 3 = -\frac{9}{4} + \frac{9}{2} + 3 = \frac{9}{4} + 3 = \frac{21}{4} \]Therefore, the local maximum is \( g(-\frac{3}{2}) = \frac{21}{4} \).

Step 5: Absolute Extreme Values

To check for absolute extreme values, we compare values at the endpoints of the domain and critical points. Here, the function has no bounded endpoints (extends to infinity), so we check the behavior at infinity:- As \( t \to -\infty \), \( g(t) \to -\infty \).- As \( t \to +\infty \), \( g(t) \to -\infty \).This means \( g(t) \) has no absolute maximum or minimum within those intervals.

Step 6: Graph Verification

Using a graphing calculator or plotting software, graph \( g(t) = -t^2 - 3t + 3 \). Confirm that the function increases on \((-fty, -\frac{3}{2})\) and decreases on \((-\frac{3}{2}, fty)\). Also, verify the local maximum at \( t = -\frac{3}{2} \) is \( \frac{21}{4} \).

Key Concepts

Derivative

In calculus, the derivative of a function helps us understand how the function changes. This is crucial when determining the behavior of a function over an interval for insights such as whether it's increasing or decreasing. To find the derivative of a given function, we perform differentiation, which is a basic calculus operation.

For example, consider the function \( g(t) = -t^2 - 3t + 3 \). Its derivative, found by applying the power rule, is \( g'(t) = -2t - 3 \). This new function, \( g'(t) \), represents the rate of change of \( g(t) \) concerning \( t \). By setting \( g'(t) \) to zero, we can find important points on the original function, such as critical points where changes in direction might happen.

The derivative essentially tells us about the slope of the tangent line to the graph of the function at any given point. If \( g'(t) > 0 \), the function is increasing at that point. Conversely, if \( g'(t) < 0 \), the function is decreasing. Understanding this concept of the derivative is vital for analyzing functions.

Critical Points

Critical points are the values of \( t \) where the derivative of a function is either zero or undefined. These points are significant because they are potential locations for local maxima or minima, or they can indicate changes in the behavior of the function.

For the function \( g(t) = -t^2 - 3t + 3 \), we find the critical point by solving the equation \( g'(t) = -2t - 3 = 0 \). Solving for \( t \), we obtain \( t = -\frac{3}{2} \).

This point is important because the behavior of \( g(t) \) on either side of \( t = -\frac{3}{2} \) will tell us whether the function has a local maximum, minimum, or neither at that point. By testing intervals around \( t = -\frac{3}{2} \), we can observe how the derivative changes and understand the overall behavior of the function at this critical point.

Increasing and Decreasing Intervals

Once we have the derivative of a function, we can determine the intervals on which the function is increasing or decreasing. This is done by choosing test points from different intervals separated by the critical points and evaluating the sign of the derivative at these points.

In the exercise, by choosing points such as \( t = -2 \) to the left of the critical point \( -\frac{3}{2} \), and \( t = 0 \) to the right, we found:

• For \( t = -2 \), \( g'(-2) = 1 \), which is positive, indicating that the function \( g(t) \) is increasing on the interval \((-\infty, -\frac{3}{2})\).
• For \( t = 0 \), \( g'(0) = -3 \), which is negative, showing that \( g(t) \) is decreasing on the interval \((-\frac{3}{2}, \infty)\).

Understanding where a function increases or decreases helps us visualize its graph and appreciate how it behaves across its domain.

Local Maximum and Minimum

Local maxima and minima are points in the function where it reaches a peak or a valley respectively, within a small neighborhood. These points are determined using the critical points and checking the sign of the derivative around these points.

In our example, at the critical point \( t = -\frac{3}{2} \), we found that the function changes from increasing to decreasing. This indicates the presence of a local maximum at this point. Calculating the function value at \( t = -\frac{3}{2} \), we get the maximum value: \[ g\left(-\frac{3}{2}\right) = \frac{21}{4} \].

Identifying these local extremes within a function helps significantly in understanding its overall shape and behavior towards its extremes, despite not knowing the function's end behavior at infinity.