Problem 35 - Thomas Calculus

Question

In Exercises \(29-36 :\) a. Identify the function's local extreme values in the given domain, and say where they are assumed. b. Which of the extreme values, if any, are absolute? c. Support your findings with a graphing calculator or computer grapher. $$ h(x)=\frac{x^{3}}{3}-2 x^{2}+4 x, \quad 0 \leq x<\infty $$

Step-by-Step Solution

Answer

Local extremum at \( x = 2 \) is neither; it has no absolute maximum, only an absolute minimum at \( x = 0 \).

Step 1: Find the First Derivative

Differentiating the function \( h(x) = \frac{x^3}{3} - 2x^2 + 4x \) with respect to \( x \) yields: \[ h'(x) = x^2 - 4x + 4 \] This will be used to find critical points as potential local extrema.

Step 2: Solve for Critical Points

Find the critical points by setting the first derivative equal to zero: \[ x^2 - 4x + 4 = 0 \]This is a quadratic equation that can be factored as: \[ (x - 2)^2 = 0 \]Solving this gives the critical point \( x = 2 \).

Step 3: Determine the Nature of the Critical Point

To determine if \( x = 2 \) is a local minimum or maximum, use the second derivative test. Calculate the second derivative: \[ h''(x) = 2x - 4 \]Evaluate the second derivative at the critical point \( x = 2 \):\[ h''(2) = 2(2) - 4 = 0 \]Since \( h''(2) = 0 \), the test is inconclusive, and further testing or graphing is needed.

Step 4: Analyze Local Extrema with Graph or Alternately Test Intervals

If graphing is allowed, use a graphing device to determine the shape of the function around \( x = 2 \). It would show a point of inflection rather than a local extremum due to double root characteristic.Alternately, manually check values of \( h'(x) \) just below and above \( x = 2 \) to see slope change or use graph evaluation in your calculator for the same conclusion.

Step 5: Check End Behavior and Absolute Extrema

Check the behavior as \( x \to \infty \). Since the leading coefficient (of \( x^3 \)) is positive, \( h(x) \to \infty \). Hence, no absolute maximum exists. Because the polynomial continues increasing, the absolute minimum must occur within the domain including the starting point, hence at \( x = 0 \).

Step 6: Evaluate Function at Specific Points

Substitute \( x = 0 \) into \( h(x) \) for initial or boundary value:\[ h(0) = \frac{0^3}{3} - 2 \cdot 0^2 + 4 \cdot 0 = 0 \]Since the function is increasing after \( x = 0 \) in direction of infinity, this is the minima by non-decreasing nature post \( x>0. \)

Key Concepts

Critical Points

Critical points are vital for understanding the potential local extreme values of a function. They occur where the first derivative of a function is zero or undefined.

In this exercise, we find the critical points of the function \( h(x) = \frac{x^3}{3} - 2x^2 + 4x \) by first computing its first derivative: \( h'(x) = x^2 - 4x + 4 \).

Setting the first derivative to zero, we solve \( x^2 - 4x + 4 = 0 \), resulting in the critical point \( x = 2 \).

This tells us that at \( x = 2 \), the function may have a local maximum, a local minimum, or neither. Further analysis is needed to determine which is the case.

First Derivative

The first derivative of a function gives us vital insight into its increasing or decreasing intervals. By differentiating the given function, we have \( h'(x) = x^2 - 4x + 4 \).

This first derivative helps us locate critical points by setting it equal to zero. Doing so in this exercise yielded the critical point \( x = 2 \).

Besides locating critical points, the sign of the first derivative in different intervals affects the slope of the function, helping to identify where it is increasing or decreasing. Checking points around the critical value further helps understand the behavior of the function, particularly if graphing is not available.

Second Derivative

The second derivative of a function, denoted \( h''(x) \), is useful for determining the concavity of the graph and the nature of critical points.

For our function, we calculate \( h''(x) = 2x - 4 \). Evaluating this second derivative at our critical point \( x = 2 \) gives \( h''(2) = 0 \), making the second derivative test inconclusive.

When the second derivative is zero, it neither confirms a local minimum nor maximum but suggests a possible inflection point. In such cases, it is often helpful to examine the surrounding values or graph the function to determine the behavior around that point.

Graph Analysis

Graph analysis involves examining how the function behaves across its domain, especially around critical points. A graphical representation provides additional insight where algebraic tests might be inconclusive.

In this problem, after determining that \( x = 2 \) is a critical point with an inconclusive second derivative, graph analysis shows the function's behavior near this point.

The graph indicates a point of inflection rather than a local extrema at \( x = 2 \), supported by the polynomial's double root characteristic in the derivative.

Furthermore, analyzing the end behavior as \( x \to \infty \) and the value at the domain's boundary, \( x = 0 \), establishes \( x = 0 \) as an absolute minimum since the function continues to increase afterwards.