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Chapter 1: Problem 77

Plot the given curve in a viewing window containing the given point \(P\). Zoom in on the point \(P\) until the graph of the curve appears to be a straight line segment. Compute the slope of the line segment: It is an approximation to the slope of the curve at \(P\). $$ y=2 x /\left(x^{2}+1\right) ; P=(0,0) $$

Short Answer

Expert verified
The slope of the curve at \(P\) is approximately 2.

Step by step solution

01

Understand the Curve Equation

The given curve is described by the equation \( y = \frac{2x}{x^2 + 1} \). This is a rational function, where the numerator and denominator are polynomial functions of \(x\). Our goal is to explore the behavior of this curve around the point \(P = (0,0)\).
02

Plot the Curve in a Viewing Window

To plot the curve, choose a viewing window that includes the point \(P = (0,0)\). Suggested x-range is \([-1, 1]\) and y-range is \([-1, 1]\). Use graphing software or a graphing calculator to visualize the curve.
03

Zoom In on the Point P

Zoom in around \(P = (0,0)\), adjust the viewing window multiple times (e.g., \([-0.1, 0.1]\) for \(x\) and \([-0.1, 0.1]\) for \(y\)) to make the curve appear as a straight line at this locality.
04

Determine the Slope of the Straight Line

Once the curve appears linear near \(P\), identify two nearby points on the line segment, for example, \( (x_1, y_1)\) and \( (x_2, y_2)\). Calculate the slope \( m \) using the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
05

Verify the Slope with Calculus

To check accuracy, compute the derivative \( \frac{dy}{dx} = \frac{d}{dx}\left( \frac{2x}{x^2 + 1} \right) \) using derivative rules. Evaluate \( \frac{dy}{dx} \) at \( x = 0 \) to verify the slope. After calculations, verify the slope is 2.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Function
A rational function is a fraction where both the numerator and the denominator are polynomials. In this exercise, the given equation \( y = \frac{2x}{x^2 + 1} \) represents a rational function because the numerator, \( 2x \), and the denominator, \( x^2 + 1 \), are polynomials.
The importance of identifying a function as rational lies in understanding its characteristics:
  • Asymptotes: Rational functions can have vertical asymptotes. However, in our function, since the denominator \( x^2 + 1 \) is always positive for real numbers, there are no vertical asymptotes.
  • Behavior: They often approach certain values as \( x \to \infty \) or \( x \to -\infty \). Here, the function will approach zero, as the degree of the numerator is less than that of the denominator.
This understanding helps us predict the shape and the boundary behavior of the graph of the curve in our given task.
Slope
The concept of slope is integral to understanding how steep a line is. In calculus, we refer to the slope in terms of the derivative, especially when dealing with functions.
The geometric interpretation:
  • The slope of a line between two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated using \( m = \frac{y_2 - y_1}{x_2 - x_1} \).
  • When zoomed into a curve at a certain point so that it appears as a straight line, the slope of this line is our approximation of the slope of the function at that particular point.
This step is vital because it gives a precise measure of the rate of change of the function at point \( P \). For our given curve, after zooming in around \( (0,0) \), the line's slope gives us an approximate value of the derivative at that point.
Derivative
A derivative helps us understand the rate of change of a function. It essentially gives us the slope of the tangent line to the curve at any given point. Calculating derivatives is a foundational calculus skill.
  • The derivative of a function \( y = f(x) \) with respect to \( x \) is denoted by \( \frac{dy}{dx} \).
  • For our function \( y = \frac{2x}{x^2 + 1} \), finding the derivative involves using the quotient rule: \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{v \cdot du - u \cdot dv}{v^2} \), where \( u = 2x \) and \( v = x^2 + 1 \).
  • After computing, this result allows us to find the slope at specific points, like \( x = 0 \), to confirm our earlier graphical approximation.
By calculating \( \frac{dy}{dx} \) at \( x = 0 \), we verify that the slope of the tangential line is indeed 2, confirming earlier findings with analytical precision.
Graphing
Graphing a function provides a visual insight into its behavior over a range of values. For the exercise, graphing the curve \( y = \frac{2x}{x^2 + 1} \) helps us closely examine its behavior around the point \( P = (0,0) \).Steps to effectively graph:
  • Begin by selecting a suitable plotting window that contains the point of interest \( P \). Initially, a range from \([-1, 1]\) for both \( x \) and \( y \) was suggested.
  • Successively zoom in, narrowing the viewing window, say \([-0.1, 0.1]\) for both axes, to closely analyze the curve's behavior.
  • At deep zoom levels, portions of the curve around \(P\) should appear as a straight line, indicating the tangent and its slope.
Graphing software or graphing calculators are invaluable here. They dynamically adjust the view, allowing an iterative process to achieve the desired zoom and clarity. This approach aids in forming an intuitive feel for calculus concepts like slope and derivative at a specific point.

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Most popular questions from this chapter

A function \(f: S \rightarrow T\) is specified. Determine if \(f\) is invertible. If it is, state the formula for \(f^{-1}(t) .\) Otherwise, state whether \(f\) fails to be one-to-one, onto, or both. \(S=[-2,5), T=[-35,98), f(s)=s^{3}-27\)

Find a function \(f\) whose graph is the given curve \(\mathcal{C}\). \(\mathcal{C}\) is obtained by translating the graph of \(y=x^{3}+2 x\) down 3 units and 2 units to the right.

In Exercises \(48-52,\) describe the curve that is the graph of the given parametric equations. \(x=7, y=t^{2}+1,-1 \leq t \leq 2\)

In Exercises 93 and \(94,\) graph the inverse function of the given function \(f .\) Do not attempt to find a formula for \(f^{-1}\) \(f(x)=x^{5}-3 x^{3}+x^{2}+6 x-8,0 \leq x \leq 2\)

Graph the curves \(\mathcal{C}\) and \(\mathcal{C}^{\prime}\) in the same viewing window. \(\mathcal{C}=\left\\{(x, y): y=(x+1) /\left(x^{4}+1\right)\right\\} ; \mathcal{C}^{\prime}\) is obtained by reflecting \(\mathcal{C}\) about the origin.
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