Problem 19 - Thomas Calculus

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. $$ g(t)=-t^{2}-3 t+3 $$

Step-by-Step Solution

Answer

The function is increasing on \((-\infty, -\frac{3}{2})\) and decreasing on \((-\frac{3}{2}, \infty)\). It has a local and absolute maximum at \( t = -\frac{3}{2} \) with a value of \( \frac{21}{4} \).

Step 1: Find the Derivative

To determine where the function is increasing or decreasing, we need to find its derivative. The function given is \[ g(t) = -t^2 - 3t + 3. \] The derivative of \( g(t) \) is: \[ g'(t) = \frac{d}{dt}(-t^2 - 3t + 3) = -2t - 3. \]

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points.\[ -2t - 3 = 0 \]Solve for \( t \): \[ -2t = 3 \] \[ t = -\frac{3}{2}. \] So, the critical point is \( t = -\frac{3}{2}. \). We need to investigate intervals around this point to check for increasing or decreasing behavior.

Step 3: Test Intervals

Use the critical point to test intervals to determine where the function is increasing or decreasing. Pick test points: one in the interval \((-\infty, -\frac{3}{2})\), such as \(t = -2\), and another in the interval \((-\frac{3}{2}, \infty)\), such as \(t = 0\).For \( t = -2 \):\[ g'(-2) = -2(-2) - 3 = 4 - 3 = 1. \] Since \( g'(-2) > 0 \), the function is increasing on \((-\infty, -\frac{3}{2})\).For \( t = 0 \):\[ g'(0) = -2(0) - 3 = -3. \] Since \( g'(0) < 0 \), the function is decreasing on \((-\frac{3}{2}, \infty)\).

Step 4: Determine Local and Absolute Extremes

From the previous analysis, we know \( t = -\frac{3}{2} \) is a critical point. At this point, the behavior of the derivative changes from positive to negative, indicating a local maximum.Evaluate \( g(t) \) at \( t = -\frac{3}{2} \):\[ g\left(-\frac{3}{2}\right) = -\left(-\frac{3}{2}\right)^2 - 3\left(-\frac{3}{2}\right) + 3 = -\frac{9}{4} + \frac{9}{2} + 3. \]Calculate:\[ = \frac{-9}{4} + \frac{18}{4} + \frac{12}{4} = \frac{21}{4}. \]Thus, there is a local maximum at \( t = -\frac{3}{2} \), and the maximum value is \( \frac{21}{4} \). Since there are no endpoints in question, this is also an absolute maximum.

Key Concepts

Critical Points

To understand the key turning points in a function, we examine its critical points. These points occur where the derivative either equals zero or is undefined. For our function, \[ g(t) = -t^2 - 3t + 3, \]the derivative is\[ g'(t) = -2t - 3. \]To locate critical points, we solve \(-2t - 3 = 0\), which simplifies to \(t = -\frac{3}{2}\).

• Critical points help identify potential locations for local maxima or minima.
• They're pivotal in analyzing where a function transitions between increasing and decreasing.

Always check the intervals around these critical points to understand the behavior of your function.

Local Maximum

When analyzing our function, we find that at the critical point \(t = -\frac{3}{2}\), the derivative changes from positive to negative. This transition suggests a local maximum at this point.

• A local maximum is a point where the function reaches its highest value compared to points immediately around it.
• It is not necessarily the highest point on the entire function, just in its local vicinity.

For \(g(t)\), the local maximum value is determined by evaluating the function at this critical point:\[ g\left(-\frac{3}{2}\right) = \frac{21}{4}. \] This value showcases the peak of the function in that region.

Increasing and Decreasing Intervals

Understanding where a function is increasing or decreasing gives us insight into its overall shape. By using derivative tests, we can determine these intervals. For the derivative \( g'(t) = -2t - 3\):

• Choose a test point in \((-\infty, -\frac{3}{2})\), such as \(t = -2\). Here, \(g'(-2) = 1\), indicating that the function is increasing in this interval.
• Choose a test point in \((-\frac{3}{2}, \infty)\), such as \(t = 0\). Here, \(g'(0) = -3\), showing the function is decreasing in this interval.

In essence, the function starts by rising up until it reaches \(t = -\frac{3}{2}\), after which it descends.

Absolute Maximum

Not only is \( t = -\frac{3}{2} \) a local maximum, but given the behavior of the function and the absence of other local maxima, it is also the absolute maximum.

An absolute maximum is the highest point over the entire domain of the function. In our case:

• The calculated local maximum value, \( \frac{21}{4} \), is higher than any values before or after in its domain.
• No endpoints or other local peaks provide a higher value than this point.

The recognition of the absolute maximum aids in evaluating the function's overall height across its domain.