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Chapter 3: Problem 23

Graph each function. Be sure to label three points on the graph. $$f(x)=\frac{1}{x}$$

Short Answer

Expert verified
The graphed function includes points (1, 1), (-1, -1), and (2, 0.5) and asymptotes at x=0 and y=0.

Step by step solution

01

- Identify Key Points

To graph the function, first identify key points that can be used to plot the curve. In this case, select three values for x: x = 1, x = -1, and x = 2. Then, calculate the corresponding y-values using the function: For x = 1: \[f(1) = \frac{1}{1} = 1\]For x = -1: \[f(-1) = \frac{1}{-1} = -1\]For x = 2: \[f(2) = \frac{1}{2} = 0.5\]
02

- Plot the Points

Plot the points (1, 1), (-1, -1), and (2, 0.5) on the coordinate plane. These points will help visualize the behavior of the function.
03

- Draw the Hyperbola

Draw a smooth curve through the plotted points to represent the hyperbola of the reciprocal function. Note that the graph will have vertical and horizontal asymptotes at x=0 and y=0 respectively, due to the function approaching infinity as x approaches zero.
04

- Label the Points and Asymptotes

Finally, clearly label the points (1, 1), (-1, -1), and (2, 0.5), as well as the vertical asymptote (x=0) and the horizontal asymptote (y=0) on the graph.

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Key Concepts

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

plotting points
When graphing reciprocal functions like \(f(x) = \frac{1}{x}\), the first step is to plot key points. Start by choosing a few values for x. For this example, we select x = 1, x = -1, and x = 2. Then, we calculate the corresponding y-values using the function:
\[f(1) = \frac{1}{1} = 1 \]
\[f(-1) = \frac{1}{-1} = -1 \]
\[f(2) = \frac{1}{2} = 0.5 \]

These calculations give us the points (1, 1), (-1, -1), and (2, 0.5) to plot on the coordinate plane. Always ensure your points are accurate, as they form the foundation of your graph.
asymptotes
Asymptotes are lines that a graph approaches but never touches. For the reciprocal function \(f(x) = \frac{1}{x}\), we have two types of asymptotes:

  • Vertical Asymptote: This occurs at x = 0. As x gets closer to 0, \(f(x)\) increases or decreases without bound, making it approach infinity.
  • Horizontal Asymptote: This occurs at y = 0. As x goes to infinity, both positive and negative, \(f(x)\) gets closer to 0 but never actually reaches it.

When graphing, horizontal and vertical asymptotes are crucial because they guide the shape of the hyperbola. Clearly label x=0 and y=0 on your graph to indicate the asymptotes.
hyperbolas
A hyperbola is a type of curve formed by the graph of a reciprocal function. In the case of \(f(x) = \frac{1}{x}\), the graph is a hyperbola that has two branches, one in the first and third quadrants. These branches are mirrored along the lines y = x and y = -x.

Key characteristics of hyperbolas:
  • Shape: Smooth curves that move away from asymptotes.
  • Quadrants: The graph of \(f(x) = \frac{1}{x}\) lies in the first and third quadrants.
  • Approach: As x moves away from zero, the graph closely approaches the asymptotes but never touches them.

After plotting your points and drawing the asymptotes, draw a smooth curve through the points. Your hyperbola will extend outwards, approaching but never touching the asymptotes at x=0 and y=0.

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Most popular questions from this chapter

\(h(x)=x^{2}-2 x\) (a) Find the average rate of change from 2 to 4 . (b) Find an equation of the secant line containing \((2, h(2))\) and \((4, h(4))\)

\(h(x)=-2 x^{2}+x\) (a) Find the average rate of change from 0 to 3 . (b) Find an equation of the secant line containing \((0, h(0))\) and \((3, h(3))\)

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=2 x^{2}-3 x+1\)

Suppose \(f(x)=x^{3}+2 x^{2}-x+6\). From calculus, the Mean Value Theorem guarantees that there is at least one number in the open interval (-1,2) at which the value of the derivative of \(f\), given by \(f^{\prime}(x)=3 x^{2}+4 x-1\), is equal to the average rate of change of \(f\) on the interval. Find all such numbers \(x\) in the interval.

Find the midpoint of the line segment connecting the points (-2,1) and \(\left(\frac{3}{5},-4\right)\)
See all solutions

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