Problem 10 - 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. $$ g(t)=-3 t^{2}+9 t+5 $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, \frac{3}{2})\) and decreasing on \((\frac{3}{2}, \infty)\); local and absolute maximum is \(\frac{47}{4}\) at \(t = \frac{3}{2}\).

Step 1: Find the derivative

Calculate the first derivative of the function \( g(t) = -3t^2 + 9t + 5 \). The derivative \( g'(t) \) will help us find critical points, which we use to determine where the function is increasing or decreasing.\[ g'(t) = \frac{d}{dt}(-3t^2 + 9t + 5) = -6t + 9 \]

Step 2: Find critical points

Set the derivative equal to zero to find critical points. These are the points where the function may change from increasing to decreasing or vice versa.\[ -6t + 9 = 0 \]Solve for \( t \):\[ t = \frac{9}{6} = \frac{3}{2} \]

Step 3: Determine intervals of increase and decrease

Use the critical point from Step 2 to test intervals around \( t = \frac{3}{2} \). Test a point from each interval in the derivative to see if the function is increasing or decreasing.- For \( t < \frac{3}{2} \), choose \( t = 0 \): \( g'(0) = -6(0) + 9 = 9 > 0 \) (increasing)- For \( t > \frac{3}{2} \), choose \( t = 2 \): \( g'(2) = -6(2) + 9 = -3 < 0 \) (decreasing)

Step 4: Identify local extreme values

From our analysis of intervals, the function changes from increasing to decreasing at \( t = \frac{3}{2} \). This point is a local maximum. To find the value of the maximum, substitute \( t = \frac{3}{2} \) back into the original function:\[ g\left(\frac{3}{2}\right) = -3\left(\frac{3}{2}\right)^2 + 9\left(\frac{3}{2}\right) + 5 \]\[ = -3 \left(\frac{9}{4}\right) + \frac{27}{2} + 5 = -\frac{27}{4} + \frac{54}{4} + \frac{20}{4} \]\[ = \frac{47}{4} \]Thus, the local maximum is \( \frac{47}{4} \) at \( t = \frac{3}{2} \).

Step 5: Determine if the extreme value is absolute

Analyze the behavior of \( g(t) \) as \( t \) approaches positive and negative infinity. Since the leading coefficient of \( -3t^2 \) is negative, the parabola opens downwards, meaning it tends to negative infinity on both ends.Thus, the local maximum at \( t = \frac{3}{2} \) is also the absolute maximum of the function.

Step 6: Verify using a graphing calculator or computer grapher

Plot the function \( g(t) = -3t^2 + 9t + 5 \) using a graphing tool. Confirm the intervals of increase and decrease, as well as the location of the maximum point at \( t = \frac{3}{2} \) where \( g(t) = \frac{47}{4} \). The graph should show a downward-opening parabola with a peak at this point.

Key Concepts

Derivative

In calculus, the derivative of a function measures how a function's value changes as the input changes. It's essentially the function's rate of change. For the function given in the exercise, the derivative is calculated as:

• Calculate the difference in the function's values for small changes in the input.
• In this case, the function is a quadratic, and its derivative is a simple linear equation: \[ g'(t) = -6t + 9 \]

Derivatives are foundational tools used to identify several key features of functions, like how they increase or decrease. Knowing how to calculate and interpret derivatives is an essential skill in calculus, leading us directly to the next important topic: critical points.

Critical Points

Critical points are where the derivative of a function is zero or undefined. These points are crucial in identifying potential maxima, minima, and points of inflection. Here's how you find them:

• Set the derivative equal to zero.
• For example, \(-6t + 9 = 0\) is solved to find that the critical point is \(t = \frac{3}{2}\).
• These points are candidates for local extrema - places where the function may peak or trough.

Critical points act like road signs for a function's graph, indicating where something important might be happening. Once you have these points, you can further investigate the behavior of the function around them.

Local Extremum

A local extremum is a point in a function where it reaches a local maximum or minimum value. Identifying these points involves:

• Analyzing intervals around critical points to see where the function increases or decreases.
• In the given example, the function \(g(t)\) changes from increasing to decreasing at \(t = \frac{3}{2}\).
  • So, \(t = \frac{3}{2}\) is a local maximum point.
• Plug these points back into the original function to find their corresponding values.
• For instance, substituting \(t = \frac{3}{2}\) into \(g(t)\) gives \( \frac{47}{4} \) as the local maximum value.

While local extrema provide critical insights into a function's behavior, they also help in understanding the function's graph and potential global behavior further.

Graphing Calculator

Using a graphing calculator or computer grapher is a practical way to visualize mathematical functions. These tools can:

• Confirm the calculations done by hand, making sure your intervals of increase and decrease are correct.
• Show the location of the maximum or minimum points.
• Visualize the overall behavior of the function, such as its downward-opening parabola shape.

For the assigned problem, plotting \(g(t) = -3t^2 + 9t + 5\) on a graphing tool should reflect the findings: an absolute maximum at \(t = \frac{3}{2}\), confirming our analysis.Graphing calculators bridge the gap between analytical calculations and visualization, helping to intuitively understand complex functions.