Question

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

Answer 1

`y = 32/33 x - 4127/528 Leftrightarrow 528y = 512x - 4127`

Explanation:

`f(4) = 1/16 - 4 Rightarrow P = (4, -63/16)`

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

`(d)/(dx)f(4) = -2/4^3 - 1 = (-1-32)/32 = m` // inclination of the tangent line

`a = -1/m = 32/33` // inclination of the normal line `n`

`P in n: y = ax + b` // let's determine b

`-63/16 = 32/33 cdot 4 + b`

`b = -63/16 - 128/33 = frac{-63*33 - 128*16}{16*33} = -4127/528`