What is the equation of the line normal to  f(x)=1/(e^x+4) at  x=6?

Oct 22, 2017

$y = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right) + \left(\frac{1}{{e}^{6} + 4}\right)$

Explanation:

To find the equation of Normal to the line f(x)=1/(e^x+4 we need to find the slope of the normal and the $y$ coordinate at ${x}_{1} = 6$

We know the slope of normal is $\frac{- 1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$ so ,

$y = \frac{1}{{e}^{x} + 4}$

Differentiate both sides with respect to $x$ using the chain rule.

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{x}}{{e}^{x} + 4} ^ 2$

So the slope of the normal ( ${m}_{n}$) is $\frac{- 1}{\frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{- 1}{\frac{- {e}^{x}}{{e}^{x} + 4} ^ 2}$

Therefore ${m}_{n} = {\left({e}^{x} + 4\right)}^{2} / {e}^{x}$

Now to find the ${y}_{1}$ coordinate when ${x}_{1} = 6$

We know $f \left(x\right) = \frac{1}{{e}^{x} + 4}$

$f \left(6\right) = \frac{1}{{e}^{6} + 4}$

Therefore the ${y}_{1}$ coordinate is $\frac{1}{{e}^{6} + 4}$

We know the equation of a line is

$\left(y - {y}_{1}\right) = {m}_{n} \left(x - {x}_{1}\right)$

$\left(y - \frac{1}{{e}^{6} + 4}\right) = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right)$

Therefore $y = \left({\left({e}^{6} + 4\right)}^{2} / {e}^{6}\right) \left(x - 6\right) + \left(\frac{1}{{e}^{6} + 4}\right)$