Problem 120 - Calculus Volume 1

Question

For the following exercises, find the equation of the tangent line \(T(x)\) to the graph of the given function at the indicated point. Use a graphing calculator to graph the function and the tangent line. $$[\mathrm{T}] \mathrm{y}=\frac{2 x}{x-1} at (-1,1)$$

Step-by-Step Solution

Answer

The tangent line is \( y = -\frac{1}{2}x + \frac{3}{2} \).

Step 1: Differentiate the Function

To find the tangent line, we first need to determine the derivative of the function. The given function is \( y = \frac{2x}{x-1} \). We'll use the quotient rule, \( \frac{u}{v} \), which states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u=2x \) and \( v=x-1 \). The derivative \( u' = 2 \) and \( v' = 1 \). So, the derivative \( y' \) is computed as follows: \[ y' = \frac{(2)(x-1) - (2x)(1)}{(x-1)^2}. \] Simplifying, we get \[ y' = \frac{2x - 2 - 2x}{(x-1)^2} = \frac{-2}{(x-1)^2}. \]

Step 2: Evaluate the Derivative at the Given Point

Now that we have the derivative, we need to find the slope of the tangent line at the point \((-1, 1)\). Substitute \( x = -1 \) into the derivative: \[ y' = \frac{-2}{((-1) - 1)^2} = \frac{-2}{4} = -\frac{1}{2}. \] Hence, the slope \( m \) of the tangent line at \( x = -1 \) is \( -\frac{1}{2} \).

Step 3: Use the Point-Slope Form to Find the Tangent Line Equation

With the slope \( m = -\frac{1}{2} \) and the point \((-1, 1)\), use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the tangent line. Substituting values, we get: \( y - 1 = -\frac{1}{2}(x + 1) \). Simplifying this equation results in: \( y - 1 = -\frac{1}{2}x - \frac{1}{2} \). Solving for \( y \), we find: \( y = -\frac{1}{2}x + \frac{1}{2} + 1 \). Hence, \( y = -\frac{1}{2}x + \frac{3}{2} \).

Step 4: Graph the Function and Tangent Line

Use a graphing calculator to plot the original function \( y = \frac{2x}{x-1} \) and the tangent line \( y = -\frac{1}{2}x + \frac{3}{2} \) on the same set of axes. Verify visually that the tangent line touches the curve at the point \((-1, 1)\) and has the correct slope.

Key Concepts

Derivative

In calculus, the concept of a derivative is fundamental to understanding rates of change and slopes of graphs. The derivative of a function represents the rate at which the function's value changes as its input changes.
It's like finding the speed of a car at a specific moment, giving you an instant rate of change.
For the function given in the exercise, we have:

• Function: \( y = \frac{2x}{x-1} \)
• To find the derivative, we use the rules of differentiation.

Derivatives can be found using various rules, and in this context, the quotient rule is particularly useful. The derivative helps in understanding how the function behaves near specific points, which we use to understand the slope of the tangent line.

Quotient Rule

The quotient rule is a formula used to differentiate functions that are ratios of two other functions. It's essential when dealing with functions expressed as fractions.
The quotient rule states:

• Given two functions \( u(x) \) and \( v(x) \), the derivative of their quotient \( \left( \frac{u}{v} \right)' \) is \( \frac{u'v - uv'}{v^2} \).

This formula is particularly handy when dealing with the function from our exercise:

• \( u = 2x \) with a derivative of \( u' = 2 \)
• \( v = x-1 \) with a derivative of \( v' = 1 \)

Substituting these into the quotient rule helps us find the derivative of the function necessary to determine the slope of the tangent line. Applying this correctly is crucial for solving problems involving tangent lines.

Slope of Tangent Line

The slope of a tangent line at a given point on a function's graph is one of the most important applications of derivatives in calculus. It tells us how steep the graph is at that exact point and whether it's increasing or decreasing.
In the problem, once we find the derivative of the function, we substitute the specific point \((-1, 1)\) into the derivative to calculate this slope.

• The derivative at this point gives the slope \( m = -\frac{1}{2} \).

The slope of the tangent line is crucial because it directly informs us of the line's steepness and direction at the given point, which is essential for plotting that line accurately.

Point-Slope Form

The point-slope form is a straightforward and very useful way of writing the equation of a line. If you know a point on the line and its slope, you can find out exactly what that line looks like.
The formula for point-slope form is:

• \( y - y_1 = m(x - x_1) \)
• \( m \) is the slope of the line.
• \((x_1, y_1)\) is a point on the line.

In our case, with a point \((-1, 1)\) and a slope \( m = -\frac{1}{2} \), this form helps in converting our findings into a tangible equation. This results in the equation \( y = -\frac{1}{2}x + \frac{3}{2} \), which is the final equation for the line tangent to the curve at the specified point.
This equation is crucial as it allows the tangent line to be graphed and verified.