Problem 43 - 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. $$g(x)=x^{2}-4 x+4, \quad 1 \leq x<\infty$$

Step-by-Step Solution

Answer

Local minimum at \(x = 2\) with value \(0\). No absolute maximum, absolute minimum at \(x = 2\).

Step 1: Identify Critical Points

To find the function's local extreme values, we need to determine where the derivative equals zero or is undefined. First, let's find the derivative of the function.The function given is \(g(x) = x^2 - 4x + 4\). Its derivative is \(g'(x) = 2x - 4\). Set the derivative equal to zero and solve for \(x\): \[ 2x - 4 = 0 \] \[ 2x = 4 \] \[ x = 2 \].The critical point is \(x = 2\). Since the derivative is a polynomial, it is defined everywhere.

Step 2: Determine Nature of Critical Points

Now, determine if the critical point at \(x = 2\) is a local minimum or maximum.Use the second derivative test. The second derivative of \(g(x)\) is \(g''(x) = 2\), which is positive. This indicates that \(x = 2\) is a local minimum.

Step 3: Evaluate Function at Critical Points and Domain Endpoints

Now we evaluate the function at the critical point and endpoints of the domain (though the domain extends to infinity, edges can offer limits for an increasing function).First, evaluate the function at the critical point:\[ g(2) = (2)^2 - 4(2) + 4 = 4 - 8 + 4 = 0 \].For the domain endpoint \(x = 1\):\[ g(1) = (1)^2 - 4(1) + 4 = 1 - 4 + 4 = 1 \].As \(x\) approaches infinity, \(g(x)\) increases, since the quadratic term dominates.

Step 4: Identify Absolute Extreme Values

Since the function \(g(x)\) is a parabola opening upwards (as indicated by the positive leading coefficient) and the domain is \(1 \leq x < \infty\), there is no absolute maximum, as \(g(x)\) increases indefinitely.The absolute minimum occurs at the critical point \(x = 2\), where \(g(2) = 0\).

Step 5: Verify with Graph

To verify our findings, use a graphing calculator or computer software to plot the function \(g(x) = x^2 - 4x + 4\) over the domain \(1 \leq x < \infty\).The graph should show a vertex at \(x = 2\) with a minimum value of \(0\). As \(x\) increases beyond 2, the graph should continue to rise indefinitely, confirming the absence of a maximum.

Key Concepts

Local extrema

Local extrema refer to the points on a function where it reaches either a local minimum or a local maximum within a certain interval. To identify these, we first look for critical points, which are where the derivative of the function is either zero or undefined. These points help highlight where potential local extrema might exist.
For example, consider the function \(g(x) = x^2 - 4x + 4\). The derivative \(g'(x) = 2x - 4\) is set to zero to find critical points, leading to \(2x - 4 = 0\) or \(x = 2\). This point is crucial in understanding the behavior of our function around it.
To determine whether this critical point is a minimum or maximum, often a second derivative test is used. If the second derivative is positive at this point, it indicates a local minimum; if negative, a local maximum. Here, \(g''(x) = 2\) is positive, confirming \(x = 2\) as a local minimum for the function. Evaluating the function at this critical point gives \(g(2) = 0\), confirming the minimum value.

Absolute extrema

Absolute extrema are the highest or lowest values a function achieves over its entire domain. Finding them necessitates checking both critical points and boundaries or limits of the domain.
Considering the quadratic function \(g(x) = x^2 - 4x + 4\), which is a parabola that opens upwards, suggests the function might have an absolute minimum but no maximum in the domain \(1 \leq x < \infty\).
By evaluating at the critical point \(x = 2\), the function reaches a value of \(g(2) = 0\). As \(x\) approaches the lower domain endpoint \(x = 1\), \(g(1)\) results in 1. Because the parabola rises indefinitely beyond \(x = 2\), there exists no absolute maximum, with \(g(x)\) increasing without bound as \(x\) heads towards infinity.
Therefore, while the function achieves its absolute minimum at \(x = 2\), reaching a point of \(0\), there is no absolute maximum within the given domain.

Derivative test

The derivative test, particularly in calculus, plays a key role in identifying local extrema. There are two primary derivative tests often used: the first and second derivative tests.

• The first derivative test involves checking the sign of the derivative before and after a critical point. If the derivative changes from positive to negative, you have a local maximum. If it changes from negative to positive, it's a local minimum.
• The second derivative test is generally more straightforward. By taking the second derivative of the function, you can analyze its sign at critical points. A positive second derivative signifies a local minimum, while a negative second derivative indicates a local maximum.

In our quadratic function, the second derivative \(g''(x) = 2\) is constant and positive, directly signifying a local minimum at \(x = 2\). Using these tests makes it more systematic to find where these important points lie, helping ensure accuracy and completeness when examining features of functions over specific domains.