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Magazine Chapter 4: Problem 62
Let \(y=1 /\left(1+x^{2}\right) .\) Find the values of \(x\) for which \(y\) is increasing most rapidly or decreasing most rapidly.
Short Answer
Expert verified
The function changes most rapidly near the origin, around \( x = 0 \).
Step by step solution
01
Understand the Problem
The function given is \( y = \frac{1}{1+x^2} \). We are asked to find where this function is increasing or decreasing most rapidly. This involves calculating the derivative of the function and then determining where this derivative has maximum or minimum values.
02
Find the Derivative
First, we find the derivative of \( y \) with respect to \( x \) using the quotient rule. If \( u = 1 \) and \( v = 1 + x^2 \), then \( \frac{dy}{dx} = \frac{v\cdot u' - u \cdot v'}{v^2} \). Here, \( u' = 0 \) and \( v' = 2x \). This gives us \[ \frac{dy}{dx} = \frac{-(2x)}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2}. \]
03
Identify Critical Points
To find where the rate of change is maximized or minimized, we consider the second derivative of \( y \). Before that, we can note from the first derivative \( \frac{dy}{dx} = 0 \) when \( x = 0 \), which gives a critical point. However, maximum/minimum values for the derivative occur at extreme values of \( x \).
04
Analyze the First Derivative
Evaluate the behavior of \( \frac{dy}{dx} = \frac{-2x}{(1+x^2)^2} \) as \( x \to \pm \infty \). As \( x \to \infty \) or \( x \to -\infty \), \( \frac{dy}{dx} \to 0 \). This implies the rate of change slows down at extreme ends of \( x \).
05
Conclude with Extreme Rate Changes
The maximum rate of increase or decrease can be observed near the origin. When \( x \) is slightly positive, the derivative is negative and decreases rapidly. As \( x \) approaches 0, the magnitude of the derivative is reduced, and the function \( y \) transitions from decreasing to increasing.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
Derivatives help us understand how a function behaves by showing how it changes at any point along its curve. Essentially, the derivative of a function at a given point tells us the slope of the tangent to the function at that point. For example, if we consider the function given in the exercise, \(y = \frac{1}{1+x^2}\), and find its derivative \(\frac{dy}{dx} = \frac{-2x}{(1+x^2)^2}\), it provides insights into how \(y\) changes with respect to \(x\).
- This expression, \(\frac{-2x}{(1+x^2)^2}\), is derived using the quotient rule. The rule is specifically useful when dealing with functions that are ratios, ensuring accurate calculation of instantaneous rates of change.
- By analyzing \(\frac{dy}{dx}\), we determine how the function is increasing or decreasing at different points.
Exploring Critical Points
Critical points occur where the derivative of a function is zero or undefined. At these points, a function can change its rate of increase or decrease, leading to local minima or maxima. In the case of our function \(y = \frac{1}{1+x^2}\), the critical point is found at \(x = 0\) because here, \(\frac{dy}{dx} = 0\).
- This means at \(x = 0\), the function stops decreasing and begins increasing.
- Critical points show where a function might change direction, making them important for finding where a function's rate of change could be most extreme.
Understanding Rate of Change
The rate of change helps us measure how a quantity changes with respect to another. In calculus, it is expressed through derivatives. The rate of change at a particular point can tell us whether a function is increasing or decreasing rapidly or slowly.
- In our function \(y = \frac{1}{1+x^2}\), the rate of change is described by the derivative \(\frac{dy}{dx} = \frac{-2x}{(1+x^2)^2}\).
- As \(x\) approaches 0 from either side, the magnitude of the rate of change diminishes, explaining why near the origin, the greatest change in rate occurs.
