Question

How do you find the linearization of `f(x) = (3x-2)/x^2` at x=2?

Answer 1

`y=(-1/4)x+3/2`

Explanation:

`f'(x)=(3(x^2)-2x(3x-2))/(x^2)^2`
`f'(x)=(3x^2-6x^2+4x)/x^4`
`f'(x)=(-3x^2+4x)/x^4`
`f'(x)=(-3x+4)/x^3`

Let's compute `f'(2)`
`f'(2)=(-3*2+4)/2^3`
`f'(2)=-2/2^3`
`f'(2)=-1/4`
And
`f(2)=(3*2-2)/2^2`
`f(2)=(6-2)/4`
`f(2)=4/4=1`

The linearizarion at `x=2`
`y-1=-1/4(x-2)`
`y=(-1/4x)+2/4+1`
`y=(-1/4)x+3/2`