/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 In Exercises 43-50, (a) find the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 43-50, (a) find the slope of the graph of \(f\) at the given point, (b) use the result of part (a) to find an equation of the tangent line to the graph at the point, and (c) graph the function and the tangent line. \(f(x)= \dfrac{1}{x+5}, \quad (-4, 1)\)

Short Answer

Expert verified
The slope of the function at the point (-4,1) is -1. The equation of the tangent line at this point is \(y=-x-3\).

Step by step solution

01

Calculating the derivative

First, we need to find the derivative of \(f(x)= \dfrac{1}{x+5}\). Use power rule for derivatives which states that \((x^n)'=nx^{n-1}\), set \(n=-1\). The derivative is then \(f'(x)= \dfrac{-1}{(x+5)^2}\).
02

Evaluating the derivative at the point (-4,1)

To find the slope of the tangent line at the point (-4,1), we substitute x=-4 into the derivative, which yields \(f'(-4)= \dfrac{-1}{(-4+5)^2}= -1\). The slope of the tangent line at the point (-4,1) is thus -1.
03

Finding the equation of the tangent line

The equation of a line is in the form \(y=mx+c\), where m is the slope and c is the y-intercept. For a tangent line, m is the derivative at the given point and c can be found by substituting the coordinates of the point into the equation. Doing so gives: \(1=-1*(-4)+c\) implying \(c=-3\). Therefore the equation for the tangent line is \(y=-x-3\).
04

Graphing the function and the tangent line

Plot the function \(f(x)= \dfrac{1}{x+5}\) and the tangent line \(y=-x-3\). The point at which they touch is (-4,1).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Slope of a Graph
Understanding the slope of a graph is fundamental in calculus and represents the rate at which a function is increasing or decreasing at a particular point. Mathematically, slope is the ratio of the vertical change to the horizontal change between two distinct points on a line. Intuitively, if you were to walk along a graphed line, the slope tells you how steep the hill would be under your feet.

When dealing with curves, the concept of slope still applies, but it’s more dynamic as the slope may change from point to point. This is where the derivative comes into play — it measures the slope of a tangent line to the curve at any given point. In our exercise, the slope at the point (-4,1) on the graph of the function f(x) = 1/(x+5) is found using the derivative, and it turns out to be -1. This means that at this point, the graph is decreasing; for a small increase in x, the value of f(x) decreases.
Derivative of a Function
The derivative of a function at a certain point can be thought of as the function's instantaneous rate of change at that point, analogous to the velocity of an object in physics. It gives us the slope of the tangent line to the function's graph at any given point.

The process of finding a derivative is called differentiation. In our exercise example, differentiation of the function f(x) = 1/(x+5) using the power rule yields f'(x) = -1/(x+5)^2. This formula provides the slope of the graph at any point x. By evaluating this derivative at x=-4, we can identify the slope at the exact point where we want to draw the tangent line.
Equation of a Tangent Line
The equation of a tangent line to a curve at a given point reflects the line's position and orientation. A tangent line just touches the curve at that point and has the same slope as the graph at that point. The general equation for a straight line is y = mx + c, where m is the slope and c is the y-intercept.

For deriving the equation of a tangent line, we need two things: the slope from the derivative at the point of tangency and the coordinates of that point to solve for c, the y-intercept. By correctly applying these values, as shown in our book's example, we arrive at the equation y = -x - 3. This equation efficiently encapsulates how the tangent line relates spatially to the original curve at the point (-4,1).
Graphing Functions and Tangents
Graphing is a visual tool that helps us see the behavior of functions and their tangents. By plotting the original function on a coordinate plane, we can draw the tangent line at the point of interest to get a clearer picture of how they interact. A well-drawn graph illustrates the concepts of slope and derivatives in a more intuitive way.

For graphing, start with plotting the original function, taking into account its continuity and limits. Then, graph the tangent line using the point of tangency and its slope. As seen in our exercise, both the curve f(x) = 1/(x+5) and the tangent line y = -x - 3 intersect at the point (-4,1). This coincidence on the graph confirms the accuracy of our previous analytical findings, making the abstract concepts of slopes, derivatives, and equations of tangent lines much more tangible.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) = 2x - x ^3 $$ Interval \( [0, 1] \)

Use a graphing utility to graph the two functions given by $$ y_{1}=\frac{1}{\sqrt{x}} \text { and } y_{2}=\frac{1}{\sqrt[3]{x}} $$ in the same viewing window. Why does not appear to the left of the axis? How does this relate to the statement at the top of page 882 about the infinite limit $$ \lim _{x \rightarrow-\infty} \frac{1}{x^{r}} ? $$

CONJECTURE Consider the functions \(f(x) = x^3\) and \(g(x) = x^3\). (a) Sketch the graphs of \(f\) and \(f'\) on the same set of coordinate axes. (b) Sketch the graphs of \(g\) and \(g'\) on the same set of coordinate axes. (c) Identify any pattern between the functions \(f\) and \(g\) and their respective derivatives. Use the pattern to make a conjecture about \(h'(x)\) if \(h(x) = x^n\), where \(n\) is an integer and \(n \geq 2\).

PROFIT The profit \(P\) (in hundreds of dollars) that a company makes depends on the amount \(x\) (in hundreds of dollars) the company spends on advertising. The profit function is given by \(P(x) = 200 + 30x - 0.5x^2\). Using your knowledge of the slopes of tangent lines, show that the profit is increasing on the interval \([0, 20]\) and decreasing on the interval \([40, 60]\).

NUMERICAL AND GRAPHICAL ANALYSIS In Exercises 35-38, (a) complete the table and numerically estimate the limit as \(x\) approaches infinity, and (b) use a graphing utility to graph the function and estimate the limit graphically. $$ f(x) = 4(4x - \sqrt{16x^2 - x}) $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.