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

Sep 27, 2016

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:

$f ' \left(x\right) = \frac{{e}^{- 2 x} \left({e}^{x} - 1\right)}{2 \sqrt{{e}^{x} - x + 1}} - 2 {e}^{2 x} \sqrt{{e}^{x} - x + 1}$

Evaluate at x = 0:

$f ' \left(0\right) = \frac{{e}^{- 2 \left(0\right)} \left({e}^{0} - 1\right)}{2 \sqrt{{e}^{0} - 0 + 1}} - 2 {e}^{2 \left(0\right)} \sqrt{{e}^{0} - 0 + 1}$

$f ' \left(0\right) = \frac{\left(1\right) \left(1 - 1\right)}{2 \sqrt{1 + 1}} - \sqrt{1 + 1}$

$f ' \left(0\right) = - \sqrt{2}$

The slope of the normal line is:

$- \frac{1}{-} \sqrt{2} = \frac{\sqrt{2}}{2}$

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

$f \left(0\right) = \sqrt{2}$

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

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