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

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Answer 1 · 2015-11-24T03:46:56

#y=-x+3#

Use #f'(x)# to find the slope of the tangent line.

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

Using the Chain Rule:

#f'(x)=-(x-1)^-2stackrel"this = 1"overbrace(d/dx[x-1]#

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

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

We can use point-slope form to write the equation of the tangent line with slope #-1# at the point on #f#, #(2,1)#.

#y-1=-(x-2)#

In slope-intercept form: #y=-x+3#