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

1 Answer
Dec 12, 2016

The equation of the tangent line is:

y(x) = 2/9x -1/9

Explanation:

The equation of a line tangent to the curve y=f(x) in x = bar x is given by:

y(x) = f(barx)+f'(barx)(x-barx)

We have:

f(x) =1/(x^2-4)

f'(x) = (-2x)/((x^2-4)^2)

For barx =-1

f(-1) =1/((-1)^2-4) = -1/3

f'(-1) = (-2(-1))/(((-1)^2-4)^2)=2/9

So that the equation of the tangent line is:

y(x) = -1/3+2/9(x+1) = 2/9x -1/9

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