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Find equations of the tangent lines to the curve \[ y=\frac{x-1}{x+1} \] that are parallel to the line \( x-2 y=4 \). (Enter your answers as a comma-separated list of equations.) F

Ask by Washington Bradley. in the United States
Mar 15,2025

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Answer

The tangent lines parallel to \( x - 2y = 4 \) are: \[ y = \frac{1}{2}x - \frac{1}{2}, \quad y = \frac{1}{2}x + \frac{7}{2}. \]

Solution

We are given the curve \[ y=\frac{x-1}{x+1} \] and we need to find the equations of the tangent lines to this curve that are parallel to the line \[ x-2y=4. \] --- **Step 1. Find the slope of the given line.** Rewrite the line \( x-2y=4 \) in slope-intercept form: \[ x-2y=4 \implies -2y = 4-x \implies y=\frac{1}{2}x-2. \] Thus, the slope of the line is \[ m=\frac{1}{2}. \] --- **Step 2. Differentiate the curve.** The curve is given by \[ y=\frac{x-1}{x+1}. \] Differentiate using the quotient rule: \[ y' = \frac{(x+1)\cdot(1) - (x-1)\cdot(1)}{(x+1)^2} = \frac{x+1-x+1}{(x+1)^2} = \frac{2}{(x+1)^2}. \] --- **Step 3. Find the x-coordinates where the slope is \(\frac{1}{2}\).** Set the derivative equal to \(\frac{1}{2}\): \[ \frac{2}{(x+1)^2} = \frac{1}{2}. \] Cross-multiply: \[ 2\cdot2 = (x+1)^2 \implies (x+1)^2 = 4. \] Take the square root: \[ x+1 = \pm2. \] Solving for \(x\): - If \(x+1 = 2\), then \(x = 1\). - If \(x+1 = -2\), then \(x = -3\). --- **Step 4. Find the corresponding y-coordinates on the curve.** For \(x=1\): \[ y=\frac{1-1}{1+1}=\frac{0}{2}=0. \] For \(x=-3\): \[ y=\frac{-3-1}{-3+1}=\frac{-4}{-2}=2. \] Thus, the points of tangency are: \[ (1,0) \quad \text{and} \quad (-3,2). \] --- **Step 5. Find the equations of the tangent lines.** Using the point-slope form \( y-y_0=m(x-x_0) \) with \( m=\frac{1}{2} \): - For point \( (1,0) \): \[ y-0=\frac{1}{2}(x-1) \implies y=\frac{1}{2}x-\frac{1}{2}. \] - For point \( (-3,2) \): \[ y-2=\frac{1}{2}(x+3) \implies y=\frac{1}{2}x+\frac{3}{2}+2=\frac{1}{2}x+\frac{7}{2}. \] --- **Final Answer** \[ y=\frac{1}{2}x-\frac{1}{2}, \quad y=\frac{1}{2}x+\frac{7}{2}. \]

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To find the tangent lines to the curve \( y=\frac{x-1}{x+1} \) that are parallel to the line \( x - 2y = 4 \), we first need to determine the slope of the given line. Rewriting it in slope-intercept form gives us: \[ 2y = x - 4 \] \[ y = \frac{1}{2}x - 2 \] Thus, the slope \( m \) is \( \frac{1}{2} \). Next, we need to find the derivative of the curve to determine the points where the slope of the tangent line equals \( \frac{1}{2} \). The derivative \( y' \) will give us the slope of the tangent line at any point on the curve. The derivative of \( y = \frac{x-1}{x+1} \) is obtained using the quotient rule: \[ y' = \frac{(x+1)(1) - (x-1)(1)}{(x+1)^2} = \frac{(x + 1) - (x - 1)}{(x + 1)^2} = \frac{2}{(x + 1)^2} \] Setting the derivative equal to the desired slope: \[ \frac{2}{(x + 1)^2} = \frac{1}{2} \] Cross-multiplying gives: \[ 2 \cdot 2 = (x + 1)^2 \] \[ 4 = (x + 1)^2 \] Taking the square root of both sides gives two solutions: \[ x + 1 = 2 \quad \text{or} \quad x + 1 = -2 \] Thus, 1. \( x = 1 \) leads to \( y = \frac{1-1}{1+1} = 0 \) 2. \( x = -3 \) leads to \( y = \frac{-3-1}{-3+1} = 2 \) Now we can write the equations of the tangent lines at these points, using point-slope form \( y - y_1 = m(x - x_1) \): 1. For the point \( (1, 0) \): \[ y - 0 = \frac{1}{2}(x - 1) \rightarrow y = \frac{1}{2}x - \frac{1}{2} \] 2. For the point \( (-3, 2) \): \[ y - 2 = \frac{1}{2}(x + 3) \rightarrow y = \frac{1}{2}x + \frac{1}{2} \] Thus, the equations of the tangent lines are: \[ y = \frac{1}{2}x - \frac{1}{2}, \quad y = \frac{1}{2}x + \frac{1}{2} \]

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