# What is the equation of the normal line of f(x)=(x+e^(1/x))^(3/2)-x at x = 4?

Jan 1, 2017

$y = - 0.4603 x + 9.9886$. See the normal-inclusive Socratic graph.

#### Explanation:

f(4)=8.1454), nearly. The point is P(4, 8.1464).

$f ' = \frac{3}{2} \frac{1 - \frac{1}{x} ^ 2 {e}^{\frac{1}{x}}}{\sqrt{x + {e}^{\frac{1}{x}}}} - 1 = 2.17134$, nearly, at P.

The slope of the normal at P $= - \frac{1}{f ' \left(4\right)} = - .4604$, nearly.

So, the equation to the normal at p(4, 8.1464) is

$y - 8.1464 = - .4603 \left(x - 4\right)$, giving

$y = - 0.4603 x + 9.9886$

graph{(y+.46x-9.99)(y+x-(x+e^(1/x))^1.5)=0 [-36.5, 27.1, -10.88, 20.95]}