Question

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

Answer 1

• 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 `x= -1`
`y=f(-1)=e^(-1)+1=1.37`

• `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`

Answer 2

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

`~~ 0.731 x + 2.099`

Explanation:

The normal line passes through the point `(-1,f(-1))`.

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

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

`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 `y=f(x)`.