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Chapter 1: Problem 16

a) Graph the function. b) Draw tangent lines to the graph at points whose \(x\) -coordinates are \(-2,0,\) and 1 c) Find \(f^{\prime}(x)\) by determining \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\). d) Find \(f^{\prime}(-2), f^{\prime}(0),\) and \(f^{\prime}(1) .\) These slopes should match those of the lines you drew in part (b). $$f(x)=\frac{2}{x}$$

Short Answer

Expert verified
Graph \( f(x) = \frac{2}{x} \), find tangent slopes at \(-2, 0, 1\); they are \(-\frac{1}{2},\) undef., and \(-2\).

Step by step solution

01

Graph the function

To graph the function \( f(x) = \frac{2}{x} \), plot points using selected \(x\)-values to ensure you capture the behavior around the vertical asymptote at \(x = 0\) and the hyperbolic shape. Remember that the function is undefined at \(x = 0\) and exhibits two branches in the first and third quadrants.
02

Draw tangent lines

Identify the points on the curve where \(x = -2, 0,\) and \(1\). Draw tangent lines at these points by estimating the slope visually or using a graphing tool. The lines should touch the graph at exactly one point at the given \(x\)-values.
03

Use the definition of a derivative

The derivative \(f'(x)\) of a function \(f(x)\) can be found using the limit definition: \[f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}.\] Substituting \(f(x) = \frac{2}{x}\), compute: \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h}.\] Simplify the expression to find the derivative.
04

Simplify to find the derivative

Continuing from the previous step, simplify: \[f'(x) = \lim_{h \rightarrow 0} \frac{\frac{2(x) - 2(x+h)}{x(x+h)}}{h} = \lim_{h \rightarrow 0} \frac{-2h}{hx(x+h)}.\] Cancel \(h\) from the numerator and denominator to get: \[f'(x) = \lim_{h \rightarrow 0} \frac{-2}{x(x+h)}.\] As \(h \rightarrow 0\), the limit simplifies to: \[f'(x) = -\frac{2}{x^2}.\]
05

Calculate specific derivative values

Substitute \(x = -2, 0,\) and \(1\) into \(f'(x) = -\frac{2}{x^2}\) to find \(f'(-2), f'(0),\) and \(f'(1)\). For \(x = -2\), \[f'(-2) = -\frac{2}{(-2)^2} = -\frac{2}{4} = -\frac{1}{2}.\] For \(x = 0\), \(f'(0)\) is undefined as the original function is not defined at \(x = 0\).For \(x = 1\), \[f'(1) = -\frac{2}{(1)^2} = -2.\] Note that \(f'(0)\) cannot be found graphically either because the tangent line does not exist at \(x = 0\).
06

Match the slopes with tangent lines

Check that the slopes of the tangent lines drawn in Part (b) match the derivative values calculated in Step 5. Ensure that at \(x = -2\), the slope is \(-\frac{1}{2}\), and at \(x = 1\), the slope is \(-2\). The slope at \(x = 0\) cannot be matched as both the function and the derivative are undefined there.

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Key Concepts

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

Function Graphing
In calculus, graphing a function helps visualize its behavior, especially around points of interest like asymptotes or regions where the function is undefined. The function given, \( f(x) = \frac{2}{x} \), is a simple rational function. This function is undefined at \( x = 0 \) as division by zero is not permissible. When graphing, you would typically plot key points to capture the function's behavior accurately. Here:
  • The graph has a vertical asymptote at \( x = 0 \), meaning the graph approaches infinity as \( x \) approaches zero from either side.
  • The function has two hyperbolic branches, located in the first and third quadrants of the Cartesian plane, indicating where \( x \) takes positive and negative values respectively.
  • As \( x \) moves away from zero, towards positive or negative infinity, the function's value approaches zero, showing the horizontal asymptote along the \( y \)-axis.
For students, plotting a few values and drawing the smooth curve can aid in understanding this behavior.
Tangent Lines
A tangent line touches the curve of a function at precisely one point, providing the instant slope or direction at that point on the graph. Drawing tangent lines can help illustrate how a function behaves locally.
  • For our function, \( f(x) = \frac{2}{x} \), tangent lines are drawn at \( x = -2, 0, \) and \( 1 \).
  • At \( x = -2 \) and \( x = 1 \), you can visually estimate the slope by drawing a line that barely skims the curve at that exact point and extends outwards following the curve's direction.
  • At \( x = 0 \), the function is not defined, and therefore, no tangent line exists.
After sketching these lines, you'll notice that the tangent lines convey how steep or flat the curve is at those special points. This visual estimation can later be confirmed by calculating the derivative.
Limit Definition of Derivative
The limit definition of a derivative is foundational in calculus. It expresses the derivative at a point in terms of limits, which helps describe how the function changes at a very small scale.For any function \( f(x) \), the typical formula for the derivative is:\[ f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \]This formula gives us the slope of the tangent line or the rate of change of the function at a point \( x \).By substituting \( f(x) = \frac{2}{x} \) into this formula:
  • You compute \( \lim_{h \rightarrow 0} \frac{\frac{2}{x+h} - \frac{2}{x}}{h} \) and simplify.
  • By simplifying, the limit calculation will eventually give us the result as \( f'(x) = -\frac{2}{x^2} \).
  • This represents how the function changes in value instantaneously at any point, \( x \), not at \( x = 0 \) due to undefined nature.
This limit concept is a core principle that leads to the full description of how a function behaves at microscopic levels.
Slope of Tangent Line
The slope of a tangent line to a function at a particular point indicates the rate at which the y-value of the function is changing with respect to x. It is the value of the derivative at that point. For the function \( f(x) = \frac{2}{x} \), the derivative is given by:\[ f'(x) = -\frac{2}{x^2} \]Calculating specific slopes:
  • At \( x = -2 \), \( f'(-2) = -\frac{1}{2} \). This negative slope suggests that the graph is decreasing at this point, and the tangent line dips downwards to the right.
  • At \( x = 1 \), \( f'(1) = -2 \). Again, a negative slope indicates a descending graph, with even greater steepness than at \( x = -2 \).
  • Notably, \( f'(0) \) is undefined because the original function does not exist at \( x = 0 \).
These calculated slopes should correspond to the visual slopes approximated by the tangent lines drawn during the graphing process. Understanding these slopes provides insight into the nature of the function's growth or decay at those points.

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Most popular questions from this chapter

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-3 $$

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\sqrt{7 x} $$

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\sqrt[3]{8 x} $$

The cost of sending a large envelope via U.S. first-class mail in 2014 was \(\$ 0.98\) for the first ounce and \(\$ 0.21\) for each additional ounce (or fraction thereof). (Source: www.usps.com.) If \(x\) represents the weight of a large envelope, in ounces, then \(p(x)\) is the cost of mailing it, where $$ \begin{array}{l} p(x)=\$ 0.98, \text { if } \quad 0 < x \leq 1, \\ p(x)=\$ 1.19, \text { if } \quad 1 < x \leq 2, \\ p(x)=\$ 1.40, \text { if } 2 < x \leq 3, \end{array} $$ and so on, up through 13 ounces. The graph of \(p\) is shown below. Using the graph of the postage function, find each of the following limits, if it exists. $$ \lim _{x \rightarrow 3} p(x) $$
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