Question

a. Find the critical points of the following functions on the domain or on the given interval. b. Use a graphing utility to determine whether each critical point corresponds to a local maximum, local minimum, or neither. $$f(x)=x^{2}-2 \ln \left(x^{2}+1\right)$$

Short answer

Answer: The critical points are x = -1 and x = 1. At x = -1, there is a local maximum, and at x = 1, there is a local minimum.

Step-by-step solution

Find the first derivative of the given function

To find the first derivative of the function, we need to find the derivative of each term separately and then simplify the result. The given function is: $$f(x)=x^{2}-2 \ln \left(x^{2}+1\right)$$ Now, let's find the first derivative with respect to x: $$f'(x)=\frac{d}{dx}\left(x^{2}-2 \ln \left(x^{2}+1\right)\right)$$ Applying the sum/difference rule and product rule: $$f'(x)=\frac{d}{dx}\left(x^{2}\right)-2\frac{d}{dx}\left(\ln \left(x^{2}+1\right)\right)$$ Taking the derivatives: $$f'(x)=2x - 2\frac{d}{dx}\left(\ln \left(x^{2}+1\right)\right)$$ Now, we will use the chain rule for the logarithm term: $$f'(x)= 2x - 2\frac{d}{dx}(\ln u), \text{ where } u = x^{2} + 1\\ \frac{du}{dx}=2x$$ So, $$f'(x)=2x - 2\frac{1}{u}\frac{du}{dx}= 2x - 2\frac{1}{x^2+1}\cdot 2x$$

Find the critical points

Critical points occur where the derivative of a function is either zero or undefined. So let's set the first derivative equal to zero and solve for x: $$0=2x - 2\frac{1}{x^2+1}\cdot 2x$$ Now, let's try to solve for x: $$2x = \frac{4x}{x^2+1}$$ $$x^2 + 1 = 2$$ $$x^2 = 1$$ So, x = ±1. Our critical points are x = -1 and x = 1.

Determine the type of extremum using a graphing utility

To determine whether each critical point corresponds to a local maximum, local minimum, or neither, we can use a graphing utility to plot the function and observe the behavior around the critical points. Upon plotting the function: At x = -1, the graph shows that the function has a local maximum. At x = 1, the graph shows that the function has a local minimum. Therefore, at x = -1, we have a local maximum, and at x = 1, we have a local minimum.

Key concepts

First Derivative Test

The first derivative test is a crucial concept in calculus used to determine whether a function's critical points are a local maximum, a local minimum, or neither. This is accomplished by analyzing the sign changes of the first derivative before and after the critical point.
The process is straightforward: calculate the first derivative of the function and find where it is equal to zero or does not exist. These points are your critical points. After identifying these points, investigate the sign of the derivative before and after each point. If the derivative changes from positive to negative, you have a local maximum; if it changes from negative to positive, you've found a local minimum. No sign change indicates neither a maximum nor a minimum.
In our exercise, after finding the critical points at x = -1 and x = 1, we would typically use the first derivative test to look for these sign changes. Instead, the solution used a graphing utility to determine the nature of these points - a practical alternative in many contexts.

Local Maximum and Minimum

Understanding local maxima and minima is essential in calculus, especially when analyzing the behavior of functions. A local maximum occurs at a point where the function's value is greater than that of any nearby points, while a local minimum is a point where the function's value is lower than any nearby points.
These points give us valuable information about the function's peaks and troughs within a specific range. For practical applications, this might tell you the optimal points for profit in a business scenario or the minimum amount of material needed to construct something.
To improve the understanding of the local maximum and minimum in relation to the given problem, ensure to explain how the graphical representation of the function offers visual confirmation of the derivative analysis, a concept that reinforces how local extrema manifest on the graph.

Chain Rule

The chain rule is a powerful tool in calculus for differentiating compositions of functions. It essentially tells us how to take the derivative of a function that is nested within another function.
In mathematical terms, if you have a composite function denoted as \( f(g(x)) \) and you want to find its derivative \( f'(x) \), the chain rule states that you must multiply the derivative of the outer function by the derivative of the inner function, symbolically: \( f'(g(x)) \cdot g'(x) \).
In our example with \( f(x) = x^2 - 2 \ln(x^2 + 1) \), the chain rule was applied to differentiate the logarithmic term. Here, the inner function is \( u = x^2 + 1 \) and the outer function is \( \ln(u) \). The differentiation of \( u \) with respect to \( x \) gives us \( 2x \) (the derivative of the inner function).
A real-world metaphor for the chain rule could be the process of unpacking a box within a box: to reach the item inside the innermost box (the core function we're differentiating), we first need to open the outer box (apply the derivative of the outer function), and then open the inner box (apply the derivative of the inner function).