Question

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

Answer 1

slope of the Normal to the tangent at x=3 is `-1/f^'(3)=49/4`

Explanation:

First find `f^'(x)` and then find its value putting x= 3

`f^'(x)=d/(dx)(x^2-2x+4)^-1`
`=-(x^2-2x+4)^-2d/(dx)(x^2-2x+4)`
`=-(x^2-2x+4)^-2(d/(dx)(x^2)-2d/(dx)(x)+d/(dx)(4)))`
`=-(x^2-2x+4)^-2(2x-2)`
`:.f^'(x)=-(x^2-2x+4)^-2(2x-2)`
`:.f^'(3)=-(3^2-2*3+4)^-2(2*3-2)=-4/49`

Hence slope of the tangent at x=3 is `f^'(3)=-4/49`

slope of the Normal to the tangent at x=3 is `-1/f^'(3)=49/4`