Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the critical points and use the test of your choice to decide which critical points give a local maximum value and which give a local minimum value. What are these local maximum and minimum values? $$ g(x)=x^{4}+x^{2}+3 $$

Step-by-Step Solution

Answer

The function has a local minimum at \( x = 0 \) with a value of 3.

Step 1: Find the first derivative

To locate the critical points, we need to find the derivative of the function and set it equal to zero. The function given is \( g(x) = x^4 + x^2 + 3 \). The first derivative is found using basic differentiation rules. \[ g'(x) = \frac{d}{dx}(x^4 + x^2 + 3) = 4x^3 + 2x \]

Step 2: Set the first derivative to zero

To find the critical points, we solve \( g'(x) = 0 \).\[ 4x^3 + 2x = 0 \]Factor the equation: \[ 2x(2x^2 + 1) = 0 \]

Step 3: Solve for critical points

From the factored form \( 2x(2x^2 + 1) = 0 \), we find the solutions.- The first factor gives us \( 2x = 0 \), so \( x = 0 \).- The second factor \( 2x^2 + 1 = 0 \) has no real solutions because \( x^2 = -\frac{1}{2} \), which does not have a real solution. Thus, the critical point is \( x = 0 \).

Step 4: Use the second derivative test

The second derivative test can confirm whether a critical point is a local maximum or minimum. First, find the second derivative:\[ g''(x) = \frac{d}{dx}(4x^3 + 2x) = 12x^2 + 2 \]Evaluate the second derivative at the critical point \( x = 0 \):\[ g''(0) = 12(0)^2 + 2 = 2 \]Since \( g''(0) > 0 \), the function has a local minimum at \( x = 0 \).

Step 5: Calculate the local minimum value

The local minimum value of the function occurs at \( x = 0 \). Substitute \( x = 0 \) into the original function:\[ g(0) = (0)^4 + (0)^2 + 3 = 3 \]Therefore, the local minimum value is 3.

Key Concepts

Local Maximum and Minimum

In calculus, local maximum and minimum points are the peaks and troughs in the graph of a function in a small neighborhood around these points. To determine these points, we must first find the critical points of a given function. Critical points occur where the derivative of the function is zero or undefined. These points are potential candidates for local maxima and minima.

Once we've identified the critical points, we utilize tests like the first and second derivative tests to confirm whether these points represent a local maximum or minimum. A local maximum occurs if the function changes from increasing to decreasing, whereas a local minimum occurs if the function changes from decreasing to increasing. This change in behavior can be analyzed using derivatives, which ultimately describe how the function behaves locally around each critical point.

First Derivative Test

The first derivative test provides insight into the nature of critical points by examining the behavior of the function's derivative across those points. The test involves looking at the sign of the first derivative before and after the critical points to determine whether each point is a maximum or minimum.

For a function \( f(x) \) and its critical point at \( x = c \):

• If \( f'(x) \) changes from positive to negative at \( x = c \), this indicates a local maximum.
• If \( f'(x) \) changes from negative to positive at \( x = c \), this indicates a local minimum.
• If \( f'(x) \) does not change signs, \( x = c \) might not be a maximum or minimum but could be a point of inflection.

The first derivative test is incredibly helpful when analyzing the directionality and slope changes of a function around its critical points, providing essential details on the nature of these points.

Second Derivative Test

The second derivative test offers another method to determine the concavity of a function at its critical points. By examining the second derivative, we can infer whether a critical point is a local maximum, minimum, or neither.

For a function \( f(x) \) with a critical point at \( x = c \):

• If \( f''(c) > 0 \), the function is concave up, and \( x = c \) is a local minimum.
• If \( f''(c) < 0 \), the function is concave down, and \( x = c \) is a local maximum.
• If \( f''(c) = 0 \), the test is inconclusive, and other methods like the first derivative test or graphing may be required to make a determination.

In our exercise, since \( g''(0) = 2 > 0 \), the function \( g(x) \) is concave up at \( x = 0 \), confirming a local minimum at this point. This technique efficiently confirms the character of critical points without needing to analyze the overall sketch or plot of the function.

Differentiation Rules

Differentiation rules are foundational to calculating derivatives, necessary for identifying critical points and applying derivative tests. Several rules simplify finding derivatives, such as:

• Power Rule: For any term \( ax^n \), the derivative is \( anx^{n-1} \).
• Sum Rule: The derivative of a sum \( f(x) + g(x) \) is the sum of derivatives, \( f'(x) + g'((x) \).
• Constant Rule: The derivative of a constant is zero.

These rules were used in our exercise where we determined the first derivative of \( g(x) = x^4 + x^2 + 3 \) as \( g'(x) = 4x^3 + 2x \). This facilitated finding critical points. Understanding these rules ensures you can effectively convert any function into its rate of change, paving the way for further analysis through tests like the first and second derivative tests.