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Q15.

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Found in: Page 922

### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508

# Find the equation of the tangent line to the curve at the given point. Graph the curve and tangent line..$y=\frac{x}{x-1},\text{}at\text{}\left(2,2\right)$

The equation of the tangent line to the curve at a given point is$y=-x+4$ .

The graph of the curve and tangent line:

See the step by step solution

## Step 1. Given information.

The function here given is,

$\begin{array}{l}y=\frac{x}{x-1}\\ po\mathrm{int}=\left(2,2\right)\end{array}$

## Step 2. Formula used.

The tangent line to the curve$y=f\left(x\right)$ at the point $P\left(a,f\left(a\right)\right)$ is the line through p with slope,

$m=\underset{x\to a}{\mathrm{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a}$.

provided that this limit exists.

Equation of tangent line:

$y=m\left(x-a\right)+f\left(a\right)$

## Step 3. Finding the slope of tangent line at given point.

Let $y=\frac{x}{x-1}$, then the slope of the tangent line at $\left(2,2\right)$ is,

$\begin{array}{c}m=\underset{x\to 2}{\mathrm{lim}}\frac{f\left(x\right)-f\left(2\right)}{x-2}\\ =\underset{x\to 2}{\mathrm{lim}}\frac{\left(\frac{x}{x-1}\right)-\left(\frac{2}{2-1}\right)}{x-2}\\ =\underset{x\to 2}{\mathrm{lim}}\frac{\frac{x}{x-1}-\frac{2}{1}\cdot \frac{x-1}{x-1}}{x-2}\\ =\underset{x\to 2}{\mathrm{lim}}\frac{\frac{x-2x+2}{x-1}}{x-2}\\ =\underset{x\to 2}{\mathrm{lim}}\frac{-\overline{)\left(x-2\right)}}{\overline{)\left(x-2\right)}\left(x-1\right)}\\ =-\frac{1}{2-1}\\ =-1\end{array}$

The slope of the tangent line is -1.

## Step 4. Finding the equation of the tangent line at a given point.

Here, the slope of the tangent line is$m=-1$ . Equation of the tangent line is written as,

$\begin{array}{l}y=-1\left(x-2\right)+2\\ y=-x+2+2\\ y=-x+4\end{array}$ ,

Therefore, the equation of the tangent line is$y=-x+4$ .