Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Identify the critical points and find the maximum value and minimum value on the given interval. $$ g(x)=\frac{1}{1+x^{2}} ; I=[-3,1] $$

Step-by-Step Solution

Answer

The maximum value is 1 at \( x = 0 \), and the minimum value is \( \frac{1}{10} \) at \( x = -3 \).

Step 1: Find the derivative of the function

Begin by finding the derivative of the function \( g(x) = \frac{1}{1+x^2} \). Use the quotient rule, which states that if you have a function \( \frac{u}{v} \), its derivative is \( \frac{v \cdot u' - u \cdot v'}{v^2} \). For \( g(x) \), \( u = 1 \), \( v = 1 + x^2 \), \( u' = 0 \), and \( v' = 2x \). The derivative is:

Step 2: Simplify the derivative

Apply the quotient rule: \( g'(x) = \frac{(1 + x^2) \cdot 0 - 1 \cdot 2x}{(1 + x^2)^2} \). Simplifying this, we get \( g'(x) = \frac{-2x}{(1 + x^2)^2} \).

Step 3: Set the derivative to zero to find critical points

To find critical points, set the derivative equal to zero: \( \frac{-2x}{(1 + x^2)^2} = 0 \). This occurs when the numerator is zero, \( -2x = 0 \). Thus, the critical point is \( x = 0 \).

Step 4: Evaluate the function at the critical points and endpoints

The interval \( I = [-3, 1] \) includes the endpoints \( x = -3 \) and \( x = 1 \). Evaluate \( g(x) \) at these points and the critical point found within the interval. Calculate: - \( g(-3) = \frac{1}{1+(-3)^2} = \frac{1}{10} \)- \( g(0) = \frac{1}{1+0^2} = 1 \)- \( g(1) = \frac{1}{1+1^2} = \frac{1}{2} \)

Step 5: Determine maximum and minimum values

Compare the values obtained: \( \frac{1}{10} \), \( 1 \), and \( \frac{1}{2} \). The maximum value is \( 1 \) at \( x = 0 \), and the minimum value is \( \frac{1}{10} \) at \( x = -3 \).

Key Concepts

Derivative Calculation

When working with calculus, a key task is finding the derivative of a function. The derivative tells us how a function changes at any point along its domain. In other words, it provides the rate of change of the function, which is crucial in identifying critical points — places where the function's behavior potentially changes from increasing to decreasing, or vice versa.
To find the derivative of a specific function like \( g(x) = \frac{1}{1+x^2} \), it's common to use the **quotient rule** because the function is in the form of a fraction. Calculating derivatives like this is a fundamental skill for analyzing the behavior of functions. Once you have the derivative, setting it equal to zero helps you find these critical points.
This process can help you determine where a function may have local minimums, maximums, or points of inflection. Ultimately, this information is valuable in understanding the overall shape and features of the graph of the function.

Quotient Rule

The quotient rule is an essential tool for differentiation, especially when dealing with functions represented as one function divided by another function. It is used to find the derivative of a quotient of two functions. The rule states that if you have functions \( u \) and \( v \), the derivative of \( \frac{u}{v} \) is given by:

• \( \frac{v \cdot u' - u \cdot v'}{v^2} \)

For the function \( g(x) = \frac{1}{1+x^2} \), you consider \( u = 1 \) and \( v = 1 + x^2 \). This means \( u' = 0 \) because the derivative of a constant is zero, and \( v' = 2x \) because the derivative of \( x^2 \) is \( 2x \).
By applying the quotient rule, we simplify to find the derivative \( g'(x) = \frac{-2x}{(1+x^2)^2} \). This formula is then used to explore when the function's slope is zero, which occurs at the critical points.

Function Evaluation

After determining the critical points by finding where the derivative equals zero or does not exist, it's important to evaluate the original function at these points. Function evaluation is about substituting specific values into a function to understand its behavior at those points.
In the example with \( g(x) \), once the critical point \( x = 0 \) was found, the function was evaluated at this point, as well as at the endpoints \( x = -3 \) and \( x = 1 \) of the interval \([-3, 1]\).

• At \( x = 0 \), \( g(0) = \frac{1}{1 + 0^2} = 1 \).
• At \( x = -3 \), \( g(-3) = \frac{1}{1 + (-3)^2} = \frac{1}{10} \).
• At \( x = 1 \), \( g(1) = \frac{1}{1 + 1^2} = \frac{1}{2} \).

These evaluations are crucial for comparing function values and determining minimum and maximum values within the given interval.

Intervals in Mathematics

Intervals describe a range of numbers, and understanding them is crucial in calculus, especially when finding critical points and evaluating functions. An interval in mathematics, such as \([-3, 1]\), denotes all real numbers \( x \) that satisfy \(-3 \le x \le 1 \).
In this particular exercise, the interval \([-3, 1]\) helps to define where the function's critical points, minimum, and maximum need to be checked. It frames the boundary within which we analyze the function's behavior.
Working within a specified interval ensures that we only consider values relevant to a particular problem, such as determining the maximum and minimum values of \( g(x) = \frac{1}{1+x^2} \). By evaluating the function at both the critical points within the interval and the endpoints themselves, we can confidently identify the highest and lowest values of the function on this domain. This method highlights the importance of intervals in ensuring that solutions are comprehensive and focused within the desired domain.