Problem 31 - 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. $$ g(x)=x^{2}-4 x+4, \quad 1 \leq x<\infty $$

Step-by-Step Solution

Answer

The local and absolute minimum is 0 at \( x = 2 \).

Step 1: Identify the Derivative

To find the local extreme values, we first need to find the derivative of the function. The function given is \( g(x) = x^2 - 4x + 4 \). Differentiating with respect to \(x\), we get \( g'(x) = 2x - 4 \).

Step 2: Solve for Critical Points

Set the derivative equal to zero to find the critical points: \( 2x - 4 = 0 \). Solving for \(x\), we have \( 2x = 4 \) which means \( x = 2 \). This is within the domain \( 1 \leq x < \infty \).

Step 3: Determine Nature of Critical Point

We apply the second derivative test to determine the nature of the critical point. Find the second derivative: \( g''(x) = 2 \). Since \( g''(2) = 2 > 0 \), the function has a local minimum at \( x = 2 \).

Step 4: Evaluate the Function at the Critical Point

Calculate \( g(2) \) to find the function value at the local minimum. We have \( g(2) = 2^2 - 4(2) + 4 = 4 - 8 + 4 = 0 \). Thus, the local minimum value is \( 0 \) at \( x = 2 \).

Step 5: Determine if Minimum is Absolute

Since \( g(x) \) is a quadratic function opening upwards and \( x = 2 \) gives the minimum value in the domain \( 1 \leq x < \infty \), this local minimum is also the absolute minimum in the given domain.

Key Concepts

Critical Points

To identify critical points of a function, we need to calculate its derivative and then set the derivative equal to zero. The solutions of this equation are the critical points. These points are essential because they help determine where a function's slope changes from increasing to decreasing, or vice versa. In our example function, \( g(x) = x^2 - 4x + 4 \), the derivative is calculated as \( g'(x) = 2x - 4 \). By setting \( g'(x) = 0 \), we solve for \( x \) and find that the critical point is at \( x = 2 \).
This tells us that there might be a local extreme at this point, subject to further testing, since the function can either have a maximum, a minimum, or possibly no extreme at all depending on the nature of the point.

Quadratic Function

Quadratic functions are polynomial functions of degree two, which generally take the form \( f(x) = ax^2 + bx + c \). Such functions graph as parabolas in the Cartesian plane. Depending on the coefficient \( a \), the parabola opening direction can be either upwards (\( a > 0 \)) or downwards (\( a < 0 \)).
In our example, \( g(x) = x^2 - 4x + 4 \), the coefficient of \( x^2 \) is positive \( (a = 1)\), meaning the parabola opens upwards. This characteristic implies that any local minimum will also be the absolute minimum if the function is continued indefinitely in the positive direction of \( x \).
The shape of the quadratic function and its symmetry around its vertex make identifying extreme values particularly straightforward, especially when located within the domain specified.

Second Derivative Test

The second derivative test is a method to classify the nature of critical points. After finding a critical point by setting the first derivative to zero, we use the second derivative to determine the curvature at that point.
If the second derivative \( g''(x) \) is positive \((g''(x) > 0)\) at the critical point, the function is concave up, indicating a local minimum. Conversely, if \( g''(x) < 0 \), the function is concave down, suggesting a local maximum. Should the second derivative be zero, other tests are necessary to determine the nature of the point.
In our example, \( g''(x) = 2 \), which is a constant positive number. Evaluating this at \( x = 2 \), we confirm that \( g''(2) = 2 > 0 \), signaling a local minimum at this point.

Absolute Minimum

An absolute minimum of a function within a given domain is the lowest point across the entire range of the function. For the quadratic function \( g(x) = x^2 - 4x + 4 \), which opens upwards, the vertex provides this minimum point.
After applying the second derivative test, we identified that the local minimum is at \( x = 2 \) with a function value \( g(2) = 0 \).
Given that quadratic functions opening upwards have no additional lower values within the specified domain \( 1 \leq x < \infty \), this local minimum is also the absolute minimum. It is the smallest value that \( g(x) \) can attain in this interval.