Problem 20 - Thomas Calculus in SI Units

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. $$g(t)=-3 t^{2}+9 t+5$$

Step-by-Step Solution

Answer

a. Increasing on \((-\infty, \frac{3}{2})\), decreasing on \((\frac{3}{2}, \infty)\). b. Local and absolute maximum of 11.25 at \(t = \frac{3}{2}\). No absolute minimum.

Step 1: Find the First Derivative

First, we need to find the derivative of the function \( g(t) = -3t^2 + 9t + 5 \). The derivative is given by \( g'(t) = \frac{d}{dt} [-3t^2 + 9t + 5] = -6t + 9 \). This will help us determine where the function is increasing or decreasing.

Step 2: Determine Critical Points

Next, set the first derivative equal to zero to find the critical points. Solve \( -6t + 9 = 0 \) which gives \( t = \frac{9}{6} = \frac{3}{2} \). Therefore, the critical point is \( t = \frac{3}{2} \).

Step 3: Test Intervals for Increasing/Decreasing

Choose test points in the intervals \((-finity, \frac{3}{2})\) and \((\frac{3}{2}, finity)\) to determine the behavior of the function. Substituting \( t = 0 \) from \((-finity, \frac{3}{2})\), we find \( g'(0) = 9 > 0 \), indicating the function is increasing on \((-finity, \frac{3}{2})\). Substituting \( t = 2 \) from \((\frac{3}{2}, finity)\), \( g'(2) = -3 < 0 \), indicating the function is decreasing on \((\frac{3}{2}, finity)\).

Step 4: Determine Local Extreme Values

Evaluate the original function at the critical point to find the local extremum. Substitute \( t = \frac{3}{2} \) into \( g(t) \): \( g(\frac{3}{2}) = -3(\frac{3}{2})^2 + 9(\frac{3}{2}) + 5 = 11.25 \). Since the function changes from increasing to decreasing at \( t=\frac{3}{2} \), this point is a local maximum.

Step 5: Evaluate Absolute Extreme Values

Since \( g(t) ightarrow -finity \) as \( t ightarrow finity \) or \( t ightarrow -finity \), and no other critical points exist, the local maximum at \( t = \frac{3}{2} \) is the highest value of \( g(t) \) on its entire domain. This is the absolute maximum, \( g(\frac{3}{2}) = 11.25 \). There is no absolute minimum because the function decreases without bound as \( t \rightarrow \pm finity \).

Key Concepts

Critical Points

In calculus, a critical point is where the first derivative of a function is zero or undefined. Finding these points is crucial because they often indicate where the function is changing direction, which can help us identify peaks and valleys in the graph. For the function \( g(t) = -3t^2 + 9t + 5 \), we start by calculating the first derivative, which is \( g'(t) = -6t + 9 \). To find the critical points, we set this equation to zero: \( -6t + 9 = 0 \). Solving for \( t \), we get \( t = \frac{3}{2} \). This point is critical because it is where the slope of the tangent to the graph is zero, indicating a potential local extreme.

Increasing and Decreasing Functions

A function is increasing on an interval if the first derivative is positive over that interval, and decreasing if the derivative is negative. To determine where \( g(t) = -3t^2 + 9t + 5 \) is increasing or decreasing, we consider intervals around the critical point \( t = \frac{3}{2} \). We test points in the intervals \((-\infty, \frac{3}{2})\) and \((\frac{3}{2}, \infty)\).

• Substituting \( t = 0 \) in the first derivative \( g'(t) = -6t + 9 \) gives \( g'(0) = 9 \), which is positive, indicating that the function is increasing on \((-\infty, \frac{3}{2})\).
• Substituting \( t = 2 \) gives \( g'(2) = -3 \), which is negative, indicating the function is decreasing on \((\frac{3}{2}, \infty)\).

These results show the function behaves differently on each side of the critical point.

Local and Absolute Extrema

Local extrema of a function are the highest or lowest values it takes in a nearby region. They occur at critical points where the function changes direction. For \( g(t) = -3t^2 + 9t + 5 \), we determined earlier that \( t = \frac{3}{2} \) is a critical point. Evaluating the function at this point, \( g(\frac{3}{2}) = 11.25 \), we find a local maximum because the function changes from increasing to decreasing.
Since \( g(t) \) tends to \(-\infty\) as \( t \) approaches \(\pm\infty\), there is no absolute minimum; the function continuously decreases. However, the local maximum at \( t = \frac{3}{2} \) is also the absolute maximum, so \( g(\frac{3}{2}) = 11.25 \) is the highest value that \( g(t) \) reaches for all real values of \( t \). This analysis helps in understanding the overall shape of the function's graph.