Problem 19 - Applied Calculus

Question

Find all critical points and then use the first-derivative test to determine local maxima and minima. Check your answer by graphing. $$f(x)=\frac{x}{x^{2}+1}$$

Step-by-Step Solution

Answer

Critical points at \( x = \pm 1 \); local maximum at \( x = -1 \), local minimum at \( x = 1 \).

Step 1: Find the derivative

Using the quotient rule: \(f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}\)

Step 2: Find critical points

Set \(f'(x) = 0\): \(1 - x^2 = 0\), so \(x = \pm 1\). The denominator is always positive, so these are the only critical points.

Step 3: First derivative test

For \(x < -1\): \(1-x^2 < 0\), so \(f'(x) < 0\) (decreasing).
For \(-1 < x < 1\): \(1-x^2 > 0\), so \(f'(x) > 0\) (increasing).
For \(x > 1\): \(1-x^2 < 0\), so \(f'(x) < 0\) (decreasing).

Step 4: Classify critical points

At \(x = -1\): \(f\) changes from decreasing to increasing, so local minimum: \(f(-1) = \frac{-1}{2}\).
At \(x = 1\): \(f\) changes from increasing to decreasing, so local maximum: \(f(1) = \frac{1}{2}\).

Key Concepts

First-Derivative Test

The First-Derivative Test is a handy tool in calculus that helps us identify where functions have local maxima or minima. Basically, you'll need to find where the derivative of your function changes sign. Here's a simple way to think of it:

• If the derivative changes from positive to negative, the function has a local maximum at that point because the function goes up and then down.
• If the derivative changes from negative to positive, the function has a local minimum because the function goes down and then up.
• If there's no change in sign, it means there is no local extremum at that critical point.

To effectively use the first-derivative test, identify all critical points, which occur where the derivative is zero or undefined. Then, check the sign of the derivative before and after each critical point. In our example, after computing the derivative as \(f'(x) = \frac{1 - x^2}{(x^2 + 1)^2}\), we set it equal to zero and solve for \(x\). This helps us find potential locations of maxima or minima.
By applying this test, we can visualize how the function behaves around these points and confirm the existence of a local maximum, minimum, or neither.

Quotient Rule for Differentiation

The Quotient Rule is essential for differentiating functions that are given as one function divided by another. It's a key concept in calculus that allows you to find the derivative of a quotient, specifically when the function is expressed as \( \frac{u}{v} \). The rule is given by: \[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]where \( u \) and \( v \) are functions of \( x \), and \( u' \) and \( v' \) are their respective derivatives.
In the exercise concerning the function \(f(x) = \frac{x}{x^2 + 1}\), we assign \( u = x \) and \( v = x^2 + 1 \). First, compute their derivatives: \( u' = 1 \) and \( v' = 2x \). Then plug into the quotient rule formula:\[ f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2+1)^2} \]This expression gives us the derivative of the function, which is crucial when analyzing the behavior of the function, specifically finding the critical points.

Local Maxima and Minima

In calculus, local maxima and minima are points on the graph of a function where the function takes a maximum or minimum value within a certain region. These concepts are important because they help describe the points where the function changes direction.
To find local maxima and minima, you must determine the critical points by setting the derivative of the function equal to zero or identifying where it is undefined. Once critical points are found, use methods such as the first-derivative test to know if they are indeed maxima or minima. For our function, critical points occur where \(f'(x) = 0\) or where the derivative does not exist. Solving \( \frac{1 - x^2}{(x^2 + 1)^2} = 0 \) leads to determining that \( x = \pm 1 \) are critical points.

• Check behavior around these points to confirm their status as local extremum.
• Graphing the function can provide additional insights and visual confirmation of the nature of these points.

Understanding local maxima and minima enables a deeper insight into the behavior of functions, illustrating how they grow and decline locally on their respective intervals.