Question

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

Answer 1

The Normal Line is

`y=x/12-49/6` or `x-12y=98`

Explanation:

the given: `f(x) =(x-2)^(3/2)-x^3` and `x=2`

`f(2) =(2-2)^(3/2)-(2)^3`

`f(2)= -8`

the point on the curve : `(2, -8)`

compute slope `m` then use `-1/m` for the normal line

`f' (x) = (3/2)(x-2)^(1/2)-3x^2`

`f' (2) = (3/2)(2-2)^(1/2)-3(2)^2`

`f' (2) = -12`

The slope to be used to find the normal line is `-1/m`

`-1/m=-1/-12=1/12`

Determine now the Normal Line using

`y-y_1=(1/12)(x-x_1)`

`y-(-8)=(1/12)(x-2)`

after simplification the final answer:

`x-12y=98` or `y=x/12-49/6`