Question

What is the equation of the normal line of `f(x)=sqrt(e^x-x+1)/e^(2x)` at `x = 0`?

Answer 1

The slope of the tangent line is the first derivative evaluated at the the given x coordinate. The slope of the normal line is the negative of the reciprocal of the slope of the tangent line.

Explanation:

Compute the first derivative:

`f'(x) = {e^(-2x)(e^x - 1)}/{2sqrt(e^x - x + 1)} - 2e^(2x)sqrt(e^x - x + 1)`

Evaluate at x = 0:

`f'(0) = {e^(-2(0))(e^0 - 1)}/{2sqrt(e^0 - 0 + 1)} - 2e^(2(0))sqrt(e^0 - 0 + 1)`

`f'(0) = {(1)(1 - 1)}/{2sqrt(1 + 1)} - sqrt(1 + 1)`

`f'(0) = -sqrt(2)`

The slope of the normal line is:

`-1/-sqrt(2) = sqrt(2)/2`

Given `x = 0` we observe that we are computing the y intercept when we evaluate f(0):

`f(0) = sqrt(2)`

Using the slope-intercept form of the equation of a line:

`y = (sqrt(2)/2)x + sqrt(2)`