Problem 28 Find the equations of the tangen... [FREE SOLUTION]

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Calculus and Its Applications

Larry J. Goldstein, David I. Schneider, Nakhle H. Asmar

$Math Studyset 91影视 Explanations$ Math

13 Edition

Chapter 4: Problem 28

Find the equations of the tangent lines to the graph of $y=\ln |x|$ at $x=1$ and $x=-1$.

Short Answer

Expert verified

The equations of the tangent lines are $y = x - 1$ at $x = 1$ and $y = -x - 1$ at $x = -1$.

Step by step solution

Find the derivative of the function

The derivative of the function gives the slope of the tangent line. The function is given by $y = \ln |x|$. The derivative of $y = \ln |x|$ with respect to $x$ is $y' = \frac{1}{x}$.

Evaluate the derivative at the given points

Substitute $x = 1$ and $x = -1$ into the derivative to find the slopes of the tangent lines. For $x = 1$, the slope is $y'(1) = \frac{1}{1}=1$. For $x = -1$, the slope is $y'(-1) = \frac{1}{-1}=-1$.

Find the y-coordinates of the points of tangency

Substitute $x = 1$ and $x = -1$ into the original function to find the y-coordinates. For $x = 1$, $y(1) = \ln |1| = 0$. For $x = -1$, $y(-1) = \ln |-1| = 0$.

Use the point-slope form of the equation of a line

The point-slope form of the equation of a line is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. For $x = 1$, with point $(1, 0)$ and slope $1$, the equation is $y - 0 = 1(x - 1)$, simplifying to $y = x - 1$. For $x = -1$, with point $(-1, 0)$ and slope $-1$, the equation is $y - 0 = -1(x + 1)$, simplifying to $y = -x - 1$.

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Key Concepts

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

derivative of logarithmic function

When working with functions involving logarithms, understanding the derivative is crucial. For the given function, $y = \ln |x| $, you will first need to find its derivative. The derivative of $y = \ln |x| $ with respect to $x$ is $\frac{d}{dx} \big( \ln |x| \big) = \frac{1}{x}$.

This derivative tells you the rate of change of the function at any point $x$. Essentially, it shows how steep the function is at different points along the curve.

For instance, if $x$ is positive, the slope will be positive, and if $x$ is negative, the slope will be negative. This is directly seen in our tangent line slopes as we evaluate the derivative at specific points.

slope of tangent line

To find out how 'tilted' or steep a tangent line is at a given point, we refer to the slope, which can be found using the derivative.

In our example, we evaluated the derivative $\frac{1}{x}$ at the points $x=1$ and $x=-1$:

For $x = 1$, we substitute into the derivative equation to get $ \frac{1}{1} = 1 $. This slope of 1 means that for every unit increase in $x$, $y$ also increases by 1 unit.
For $x = -1$, substituting gives us $ \frac{1}{-1} = -1 $, indicating that as $x$ increases by one unit, $y $ decreases by one unit, giving us a downward slope.

Understanding the slope at specific points helps us form the equations of the tangent lines.

point-slope form equation

Once we have the slope and a specific point that the line passes through, we can use the point-slope form to write the equation of the tangent line. The general formula for the point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is our specific point.

In the given problem:

For $x = 1$:
The point is (1, 0) with slope 1. So, the equation will be $ y - 0 = 1(x - 1) $. Simplifying this, we get $ y = x - 1 $.
For $x = -1$:
The point is (-1, 0) with slope -1. The equation here is $ y - 0 = -1(x + 1) $, which simplifies to $ y = -x - 1 $.

Using the point-slope form makes it straightforward to find the linear equations of tangent lines at any given points.

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91影视

Find the equations of the tangent lines to the graph of \(y=\ln |x|\) at \(x=1\) and \(x=-1\).

Short Answer

Step by step solution

Find the derivative of the function

Evaluate the derivative at the given points

Find the y-coordinates of the points of tangency

Use the point-slope form of the equation of a line

Key Concepts

derivative of logarithmic function

slope of tangent line

point-slope form equation

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