# What is the equation of the normal line of f(x)=e^(1/x)-x at x=-1?

Feb 25, 2016
• The equation of the line is:
$y = 1.37 x + 2.7$

#### Explanation:

• to find the equation of the line, we find its slope ($m$) knowing that:
$m = f ' \left(x\right)$
• $f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left[{e}^{\frac{1}{x}} - x\right]$
by substituting $x$ with $\frac{1}{z}$ in the main equation:
$f \left(\frac{1}{z}\right) = {e}^{z} - \frac{1}{z}$
$f ' \left(x\right) = f ' \left(\frac{1}{z}\right) = \frac{d}{\mathrm{dz}} \left[{e}^{z} - \frac{1}{z}\right]$
$= {e}^{z} + \frac{1}{z} ^ 2$ (because the derivative of ${e}^{x}$ is ${e}^{x}$ )
$= {e}^{\frac{1}{x}} + {x}^{2}$

At $x = - 1$
$y = f \left(- 1\right) = {e}^{- 1} + 1 = 1.37$

• $f ' \left(- 1\right) = {e}^{- 1} + 1 = 1.37$
$\therefore m = 1.37$
• The equation of the line is:
$y - 1.37 = 1.37 \left(x + 1\right)$
$y = 1.37 x + 2.7$
Mar 2, 2016

$y = \frac{e}{e + 1} x + 2 + \frac{1}{e \left(e + 1\right)}$

$\approx 0.731 x + 2.099$

#### Explanation:

The normal line passes through the point $\left(- 1 , f \left(- 1\right)\right)$.

$f \left(- 1\right) = \frac{1}{e} + 1$

The normal line is perpendicular to the tangent line. It has a gradient of $m = \frac{- 1}{f ' \left(- 1\right)}$.

$f ' \left(x\right) = \frac{\text{d"}{"d"x}(e^{1/x}) - frac{"d"}{"d} x}{x}$

$= {e}^{\frac{1}{x}} \frac{\text{d"}{"d} x}{\frac{1}{x}} - 1$

$= - \frac{1}{x} ^ 2 {e}^{\frac{1}{x}} - 1$

$f ' \left(- 1\right) = - \frac{1}{e} - 1$

$m = \frac{- 1}{- \frac{1}{e} - 1}$

$= \frac{e}{e + 1}$

Now we know the gradient and a point, we can write the equation of the normal line in point-slope form.

$y - f \left(- 1\right) = m \left(x - \left(- 1\right)\right)$

$y - \left(\frac{1}{e} + 1\right) = \frac{e}{e + 1} \left(x + 1\right)$

Rewrite it in slope-intercept form.

$y = \frac{e}{e + 1} x + 2 + \frac{1}{e \left(e + 1\right)}$

Here is a graph of $y = f \left(x\right)$.