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

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Answer 1 · 2016-04-02T12:25:21

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

To find the full co-ordinates of the point that you're finding the tangent of, substitute #x = -3# into the original equation.

#y = f(-3) = -1/-3 = 1/3#

Which gives the #(x, y)# co-ordinates

#(-3, 1/3)#

Differentiate f(x) by switching around the function to make it more friendly and then differentiating in the normal fashion.

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

This is the gradient of the line, or #m# in the line equation #y = mx + c#. Substituting our value of #x#, #m# more properly is

#1/(-3)^2 = 1/9#, so

#y = 1/9x + c#

Substituting all of the co-ordinates into this to find #c#, we get

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

Putting this back into the form #y = mx + c#,

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

Which is the equation of the tangent.