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Q15.

Expert-verified
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$y=-x+4$ .