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

##### 1 Answer

Jan 1, 2017

#### Explanation:

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

The slope of the normal at P

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

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