Notes: Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1).

Sunday, October 27, 2013

Find the equation of the line with slope -1 that is the tangent to the curve y=1/(x-1).

To solve, let's consider the given slope of a line.

$\displaystyle{m}=-{1}$

Take note that when a line is tangent to the curve, its derivative at the point of tangency is equal to the slope.

$\displaystyle{y}'={m}$

$\displaystyle{y}'={1}$

To determine the point of tangency, take the derivative of the function of the curve.

$\displaystyle{y}=\frac{{1}}{{{x}-{1}}}$

$\displaystyle{y}={{\left({x}-{1}\right)}}^{{-{1}}}$

$\displaystyle{y}'=-{1}{{\left({x}-{1}\right)}}^{{-{1}-{1}}}\cdot{\left({x}-{1}\right)}'$

$\displaystyle{y}'=-{{\left({x}-{1}\right)}}^{{-{2}}}\cdot{\left({1}\right)}$

$\displaystyle{y}'=-\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}$

Then, plug-in the value of the derivative.

$\displaystyle-{1}=-\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle{1}=\frac{{1}}{{{\left({x}-{1}\right)}}^{{2}}}$

$\displaystyle{{\left({x}-{1}\right)}}^{{2}}={1}$

$\displaystyle{x}-{1}=\pm{1}$

$\displaystyle{x}=\pm{1}+{1}$

$\displaystyle{x}_{{1}}={0}$

$\displaystyle{x}_{{2}}={2}$

To solve for the values of y, plug-in x=0 and x=2.

$\displaystyle{x}_{{1}}={0}$

$\displaystyle{y}_{{1}}=\frac{{1}}{{{0}-{1}}}=-{1}$

$\displaystyle{x}_{{2}}={2}$

$\displaystyle{y}_{{2}}=\frac{{1}}{{{2}-{1}}}={1}$

So, there two points on the curve in which the tangent have a slope of -1. These are:

(0,-1) and (2,1).

This means that there are two lines that have a slope of -1. To determine the equation of each line, use the point-slope form.

The first line has a slope of -1 and passes the point (0,-1). So its equation is:

$\displaystyle{y}-{y}_{{1}}={m}{\left({x}-{x}_{{1}}\right)}$

$\displaystyle{y}-{\left(-{1}\right)}=-{1}{\left({x}-{0}\right)}$

$\displaystyle{y}+{1}=-{x}$

$\displaystyle{y}=-{x}-{1}$

The second line has a slope of -1 and passes the point (2,1). So its equation is:

$\displaystyle{y}-{y}_{{2}}={m}{\left({x}-{x}_{{2}}\right)}$

$\displaystyle{y}-{1}=-{1}{\left({x}-{2}\right)}$

$\displaystyle{y}-{1}=-{x}+{2}$

$\displaystyle{y}=-{x}+{3}$

Therefore, the equation of the lines with slope -1 and tangent to the given curve are:

$\displaystyle{y}=-{x}-{1}$

and

$\displaystyle{y}=-{x}+{3}$ .

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