What is the equation of the line that is normal to f(x)= x-e^sqrt(x+4)/(x-4)  at  x= 5 ?

Dec 25, 2017

y=5−e^3-6/(6+5e^3)*(x-5)

Explanation:

f(x)=x−(e^(sqrt(x+4)))/(x−4);quadquadquad $a = 5$

$y = f \left(a\right) - \frac{1}{f ' \left(a\right)} \cdot \left(x - a\right)$

$f ' \left(x\right) = 1 - \frac{{e}^{\sqrt{x + 4}} \frac{1}{2} {\left(x + 4\right)}^{- \frac{1}{2}} \left(x - 4\right) - {e}^{\sqrt{x + 4}}}{x - 4} ^ 2$

$f ' \left(5\right) = 1 - \frac{{e}^{\sqrt{5 + 4}} \frac{1}{2} {\left(5 + 4\right)}^{- \frac{1}{2}} \left(5 - 4\right) - {e}^{\sqrt{5 + 4}}}{5 - 4} ^ 2$

$f ' \left(5\right) = 1 - \frac{{e}^{3} \left(\frac{1}{2}\right) {\left(9\right)}^{- \frac{1}{2}} - {e}^{3}}{1} = 1 - \left({e}^{3} \left(\frac{1}{2 \cdot 3}\right) - {e}^{3}\right)$

$f ' \left(5\right) = 1 - {e}^{3} \left(\frac{1}{6} - 1\right) = 1 - {e}^{3} \left(- \frac{5}{6}\right) = 1 + \frac{5}{6} {e}^{3} = \frac{6 + 5 {e}^{3}}{6} \approx 17.7379$

f(5)=5−(e^(sqrt(5+4)))/(5−4)=5−e^3~~-15.085

y=5−e^3-6/(6+5e^3)*(x-5)