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Chapter 3: Problem 13

Find the slope of the function's graph at the given point. Then find an equation for the line tangent to the graph there. $$g(x)=\frac{x}{x-2}, \quad(3,3)$$

Short Answer

Expert verified
The slope at the point is -2, and the tangent line equation is \( y = -2x + 9 \).

Step by step solution

01

Find the derivative of the function

To find the slope of the tangent line at a given point, we first need the derivative of the function. Given the function \( g(x) = \frac{x}{x-2} \), we use the quotient rule for differentiation. The quotient rule states: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = x \) and \( v = x-2 \). Thus, \( u' = 1 \) and \( v' = 1 \). Substituting into the quotient rule gives: \[ g'(x) = \frac{1\cdot(x-2) - x\cdot1}{(x-2)^2} = \frac{x-2-x}{(x-2)^2} = \frac{-2}{(x-2)^2}. \]
02

Calculate the slope at the given point

With the derivative \( g'(x) = \frac{-2}{(x-2)^2} \), we find the slope at \( x = 3 \). Substitute \( x = 3 \) into \( g'(x) \): \[ g'(3) = \frac{-2}{(3-2)^2} = \frac{-2}{1} = -2. \] So, the slope of the tangent line at the point \((3, 3)\) is \(-2\).
03

Find the equation of the tangent line

Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with point \( (x_1, y_1) = (3, 3) \) and slope \( m = -2 \), we can write: \[ y - 3 = -2(x - 3). \] Simplifying, we distribute \(-2\): \[ y - 3 = -2x + 6. \] Then, add 3 to both sides to isolate \( y \): \[ y = -2x + 9. \] So, the equation of the tangent line is \( y = -2x + 9 \).

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Key Concepts

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

Derivative Calculation
The derivative is a fundamental part of calculus, allowing us to find the rate at which a function changes. Calculating derivatives is essential for understanding the behavior of functions, especially when finding tangent lines. In this problem, we have the function \( g(x) = \frac{x}{x-2} \). To find the tangent line's slope at a certain point, we need to determine the derivative, \( g'(x) \). This can show how quickly \( g(x) \) is changing at any given \( x \)-value. By doing this, we establish the slope at the exact point where we want to draw the tangent line.
Quotient Rule
When a function is the division of two other functions, its derivative can be found using the quotient rule. This rule helps us compute derivatives when dealing with fractions, like in \( g(x) = \frac{x}{x-2} \). According to the quotient rule, if a function is presented as \( \frac{u}{v} \), then its derivative is \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \). Here:
  • \( u = x \)
  • \( v = x-2 \)
  • \( u' = 1 \)
  • \( v' = 1 \)

By substituting these into the formula, we find that \( g'(x) = \frac{-2}{(x-2)^2} \). This result gives us the slope of the function at any point \( x \) on the curve.
Slope of a Function
The slope of a function at a particular point indicates how steeply the function rises or falls. It's like calculating the steepness of a hill. The derivative helps find this slope precisely. In our problem, the derivative \( g'(x) = \frac{-2}{(x-2)^2} \) tells us the function's slope at any \( x \). For the point \((3,3)\), we substitute \( x = 3 \) into \( g'(x) \) to find the slope:
\( g'(3) = \frac{-2}{1} = -2 \).
This tells us that at point \((3,3)\), the graph is going downwards with a slope of \(-2\), because it's negative. It tells us exactly how steep the tangent is at \( x = 3 \).
Point-Slope Form
Once we know the slope at a specific point, we can easily find the equation for the tangent line using the point-slope form. The point-slope form equation is given by:
\( y - y_1 = m(x - x_1) \)
where \((x_1, y_1)\) is a point on the line, and \( m \) is the slope. In our exercise, the point is \((3,3)\) and the slope is \(-2\). By substituting these values, we get:
\( y - 3 = -2(x - 3) \).
After simplifying this expression, we arrive at \( y = -2x + 9 \). This linear equation represents the tangent line to the curve at the point \((3,3)\). Thus, the point-slope form helps translate the slope and point information into a full equation for a line.

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Most popular questions from this chapter

Suppose that a piston is moving straight up and down and that its position at time \(t\) s is $$s=A \cos (2 \pi b t)$$ with \(A\) and \(b\) positive. The value of \(A\) is the amplitude of the motion, and \(b\) is the frequency (number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity, acceleration, and jerk? (Once you find out, you will know why some machinery breaks when you run it too fast.)

Find \(y^{\prime \prime}\) $$y=9 \tan \left(\frac{x}{3}\right)$$

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2$$

Find \(y^{\prime \prime}\) $$y=(1-\sqrt{x})^{-1}$$

Show that if it is possible to draw three normals from the point \((a, 0)\) to the parabola \(x=y^{2}\) shown in the accompanying diagram, then \(a\) must be greater than \(1 / 2\). One of the normals is the \(x\) -axis. For what value of \(a\) are the other two normals perpendicular?
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