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

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Answer 1 · 2017-01-10T17:43:44

The tangent line is: #y=2x-2#

The equation of the tangent line to the curve #y=f(x)# in the point #(barx,f(bar x))# is given by the formula:

# y=f(barx)+f'(barx)(x−barx)#

In our case:

#f(x)=(x^2-1)/x => f(1) = 0#

#f'(x)= frac (2x^2-(x^2-1)) (x^2) = frac (x^2+1) (x^2) => f'(1) = 2#

So the tangent line is:

#y=2(x−1)=2x−2#

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