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

##### 2 Answers

Feb 25, 2016

- The equation of the line is:

#y=1.37x+2.7#

#### Explanation:

- to find the equation of the line, we find its slope (
#m# ) knowing that:

#m=f'(x)# #f'(x)=d/dx[e^(1/x)-x]#

by substituting#x# with#1/z # in the main equation:

#f(1/z)=e^z-1/z#

#f'(x)=f'(1/z)=d/dz[e^z-1/z]#

#=e^z+1/z^2# (because the derivative of#e^x# is#e^x# )

#=e^(1/x)+x^2#

At

#f'(-1)=e^(-1)+1=1.37#

#:.m=1.37# - The equation of the line is:

#y-1.37=1.37(x+1)#

#y=1.37x+2.7#

Mar 2, 2016

#~~ 0.731 x + 2.099#

#### Explanation:

The normal line passes through the point

#f(-1) = 1/e + 1#

The normal line is perpendicular to the tangent line. It has a gradient of

#f'(x) = frac{"d"}{"d"x}(e^{1/x}) - frac{"d"}{"d"x}(x)#

#= e^{1/x}frac{"d"}{"d"x}(1/x) - 1#

#= -1/x^2 e^(1/x) - 1#

#f'(-1) = -1/e -1#

#m = frac{-1}{-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(-1) = m(x-(-1))#

#y-(1/e + 1) = frac{e}{e + 1}(x+1)#

Rewrite it in slope-intercept form.

#y = frac{e}{e+1}x + 2 + frac{1]{e(e+1)}#

Here is a graph of