Problem 42 - Thomas Calculus

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(x)=(x+1)^{2}, \quad-\infty

Step-by-Step Solution

Answer

The local and absolute minimum is 0, occurring at \( x = -1 \).

Step 1: Identify the Critical Points

To find the local extreme values of the function \( f(x) = (x+1)^2 \), we first need to find its derivative. The derivative is \( f'(x) = 2(x+1) \). To find the critical points, we set the derivative equal to zero: \( 2(x+1) = 0 \). Solving this, we get \( x = -1 \). Since the given domain is \(-\infty < x \leq 0\), \( x = -1 \) is within the domain and is a critical point.

Step 2: Determine the Nature of Critical Points

To determine whether \( x = -1 \) is a local maximum or minimum, we examine the second derivative. The second derivative is \( f''(x) = 2 \), which is positive. Since the second derivative is positive, \( x = -1 \) is a local minimum.

Step 3: Verify Boundary Values

The function is also defined up to \( x = 0 \). Evaluate \( f(x) \) at \( x = 0 \): \( f(0) = (0 + 1)^2 = 1 \). At the boundary \( x = 0 \), the function value is 1.

Step 4: Determine Absolute Extreme Values in the Domain

In the domain \(-\infty < x \leq 0\), as \( x \to -\infty \), \( f(x) \to \infty \). The lowest value of \( f \) is at the local minimum 0, which occurs at \( x = -1 \). Since the function value \( f(0) = 1 \) is higher than at \( x = -1 \), the absolute minimum value is 0 at \( x = -1 \), and no other absolute extreme values exist in the domain.

Step 5: Support Findings with a Graph

Using a graphing calculator or software, graph \( f(x) = (x+1)^2 \) on the interval \(-\infty < x \leq 0\). The graph will show a parabola opening upwards with the vertex (lowest point) at \( x = -1 \), confirming the local (and absolute) minimum at this point.

Key Concepts

Critical Points

To find local extreme values of a function, we often start by identifying critical points. Critical points occur where the derivative of the function is either zero or undefined. This is key because it helps us locate points where the function could have a local minimum or maximum.

For the function \( f(x) = (x+1)^2 \), the derivative is \( f'(x) = 2(x+1) \). By setting \( f'(x) = 0 \), we get \( 2(x+1) = 0 \). Solving, we find \( x = -1 \). Since \( x = -1 \) is within the domain \(-\infty < x \leq 0\), it is a critical point of the function.

Second Derivative Test

After finding a critical point, we use the second derivative test to determine whether it is a local minimum or maximum. The second derivative helps us understand the concavity of the function:

• If \( f''(x) > 0 \), the function is concave up, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.

The second derivative of our function \( f(x) = (x+1)^2 \) is \( f''(x) = 2 \), which is always positive. Thus at \( x = -1 \), the function is concave up, confirming a local minimum at this point.

Absolute Extreme Values

Absolute extreme values are the highest or lowest points of a function on a given domain. To find these, check both critical points and the endpoints of the domain.

For our function, \( f(x) = (x+1)^2 \), we evaluate it at the critical point and other meaningful points, like the endpoint at \( x = 0 \). We find:

• \( f(-1) = 0 \) (local minimum)
• \( f(0) = 1 \)

As \( x \to -\infty \), \( f(x) \to \infty \). Thus, the absolute minimum value is \( 0 \) at \( x = -1 \), with no absolute maximum.

Graphing Calculator

Using a graphing calculator or computational tool is invaluable for visually confirming our findings. Graph \( f(x) = (x+1)^2 \) to see the parabola shape clearly. In this graph:

• The vertex of the parabola appears at \( x = -1 \), and the parabola opens upwards.
• This supports our determination of a local and absolute minimum at \( x = -1 \).

Graphing provides an intuitive check on our analytical solutions, ensuring no mistakes were made in calculation.

Derivative

The concept of the derivative is fundamental in calculus, often describing the rate of change of a function. The first derivative gives us the slope of the function at any point, while higher order derivatives give more detailed insight into its behavior.

In this problem, the first derivative \( f'(x)=2(x+1) \) helps locate critical points and shows where the function's slope is zero. The second derivative \( f''(x) = 2 \) confirms the nature of these points through concavity. Mastering derivatives allows us to solve for optimizations and understand a function's behavior deeply.