What is the equation of the tangent line of #f(x)=-1/x# at #x=3#?

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Answer 1 · 2016-01-06T13:00:03

The equation of the tangent line is: #9y + 6 = x#.

By the power rule, we can find #f'(x)# which is the equation for the 'slope' or the derivative of #f(x)#.

#f(x) = - x^(-1)#

#f'(x) = (-1) * (-1) x^(-2)#

#f'(x) = 1/x^2#

Evaluating at the given #x# value,

#f'(3) = 1/3^2#

#f'(3) = 1/9#

The point that we're interested in is #(3, -1/3)#, where I simply evaluated #f(3)# to find the y-coordinate.

So, we know that the tangent line goes through the point #(3, -1/3)# with a gradient of #m = 1/9#.

Thus, we can find the equation of that line through the 'point-slope' method.

#y - (-1/3) = 1/9(x - 3)#

#y = 1/9 x - 3/9 - 1/3#

#y = 1/9 x - 2/3#

Which can also be written as #9y = x - 6#.