Problem 46 - Thomas Calculus in SI Units

Question

a. Identify the function's local extreme values in the given domain, and say where they occur. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$f(t)=t^{3}-3 t^{2}, \quad-\infty

Step-by-Step Solution

Answer

The local extrema are a maximum at \( t = 0 \) and a minimum at \( t = 2 \). The minimum at \( t = 2 \) is the absolute minimum. No absolute maximum exists as the maximum value at \( t = 0 \) and \( t = 3 \) are the same.

Step 1: Find the critical points

To find the critical points, first take the derivative of the function \( f(t) = t^3 - 3t^2 \). The derivative is \( f'(t) = 3t^2 - 6t \). Set \( f'(t) = 0 \) to find the critical points: \[ 3t^2 - 6t = 0 \]\[ 3t(t-2) = 0 \]Thus, the critical points are \( t = 0 \) and \( t = 2 \).

Step 2: Determine the nature of each critical point

Evaluate the second derivative \( f''(t) = 6t - 6 \) to determine the nature of the critical points:1. At \( t = 0 \), \( f''(0) = 6(0) - 6 = -6 \), indicating a local maximum.2. At \( t = 2 \), \( f''(2) = 6(2) - 6 = 6 \), indicating a local minimum.

Step 3: Evaluate endpoints in the domain

Evaluate the function at the endpoint of the domain. Since the domain is \( -\infty < t \leq 3 \), consider \( t = 3 \):\[ f(3) = 3^3 - 3(3)^2 = 27 - 27 = 0 \].

Step 4: Compare function values

Consider the function values at critical points and the endpoint:- \( f(0) = 0^3 - 3(0)^2 = 0 \)- \( f(2) = 2^3 - 3(2)^2 = 8 - 12 = -4 \)- \( f(3) = 0 \)Since the lowest value is \( f(2) = -4 \) and the highest is \( f(0) = 0 \) which is equal to \( f(3) \), the local minimum is also the absolute minimum in the given domain. There is no absolute maximum because the endpoints all give the same highest value.

Step 5: Graph the function

Plot the graph of \( f(t) = t^3 - 3t^2 \) and observe the critical points and endpoints: 1. The graph decreases to a minimum at \( t = 2 \) with the function value \( -4 \).2. It increases to \( t = 3 \) where the function value is \( 0 \), leveling off there and at \( t = 0 \). This confirms the calculated extrema.

Key Concepts

Critical Points

Critical points of a function are essential because they help us understand where a function could experience a peak, trough, or change direction. Essentially, they occur where the first derivative of the function is zero or undefined.

In our function, defined as \( f(t) = t^3 - 3t^2 \), we identify critical points by finding the derivative and setting it equal to zero. This derivative is \( f'(t) = 3t^2 - 6t \).

Setting \( f'(t) = 0 \):

• \( 3t^2 - 6t = 0 \)
• Factoring this yields \( 3t(t-2) = 0 \)
• This results in critical points at \( t = 0 \) and \( t = 2 \).

Identifying critical points is the first step towards understanding where a function may have local maxima or minima.

Derivative Test

The derivative test, mainly the first and second derivative tests, is a primary method to determine the nature of critical points.

1. **First Derivative Test**: It involves checking the sign changes around the critical points to see if a function is increasing or decreasing before and after them. In our example, we have no need to use this test because we will employ the second derivative test.

2. **Second Derivative Test**: This is more direct, as it uses the second derivative to tell if there's a concave up or down structure at critical points:

• The second derivative of our function is \( f''(t) = 6t - 6 \).
• For \( t = 0 \), \( f''(0) = -6 \), which indicates a concave down shape—meaning a local maximum.
• For \( t = 2 \), \( f''(2) = 6 \), indicating a concave up shape—meaning a local minimum.

By using the derivative test, we confirm the nature of these points, allowing us to distinguish between local minima and maxima effectively.

Graphing Functions

Graphing functions provide a visual context, aiding in the survey of a function’s behavior across a range of values. For our function, \( f(t) = t^3 - 3t^2 \), graphing helps us validate our algebraic findings.

When plotting:

• Start by identifying key features: axes intersections, critical points, and any asymptotic behavior.
• For our function, observe that the function's curve decreases towards \( t=2 \) (local minimum), then increases to \( t=3 \).
• The graph shows symmetry and slope changes at critical points.

Graphs act as a valuable tool to visually check calculations of turning points and extrema, offering a comprehensive understanding of function dynamics.

Local and Absolute Extrema

Extrema are points where a function reaches its highest or lowest values, either locally (within a particular neighborhood) or absolutely (over the whole domain).

In analyzing the function \( f(t) = t^3 - 3t^2 \):

• We located a local maximum at \( t = 0 \) (\( f(0) = 0 \)).
• There's a local minimum at \( t = 2 \) with \( f(2) = -4 \).
• Evaluating \( f(t) \) at the domain endpoint, \( t = 3 \), gives \( f(3) = 0 \), matching the local maximum value at \( t = 0 \).
• The absolute minimum is clearly at \( t = 2 \) with \( f(2) = -4 \) since it's the lowest value within the interval.

Thus, using the derivative test and graphing confirms that while there is an absolute minimum, there is no distinct absolute maximum for the given domain, as both ends of the domain have the same highest value (0).