Question

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

Answer 1

`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'= 3/2(1-1/x^2e^(1/x))/sqrt(x+e^(1/x)) -1=2.17134`, nearly, at P.

The slope of the normal at P `= -1/(f'(4))=-.4604`, nearly.

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

`y-8.1464=-.4603(x-4)`, 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]}