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

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Answer 1 · 2016-09-24T20:06:20

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

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

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

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