What is the equation of the line tangent to #f(x)=4/(x-1) # at #x=0#?

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Answer 1 · 2016-06-15T00:29:37

#y=-4x-4#

Find the point of tangency:

#f(0)=4/(0-1)=-4#

The tangent line passes through the point #(0,-4)#.

To find the slope of the tangent line, find the value of the derivative of the function at #x=0#. To differentiate the function, rewrite it:

#f(x)=4(x-1)^(-1)#

This will involve the chain rule:

#f'(x)=-4(x-1)^(-2)*d/dx(x-1)#

#f'(x)=-4/(x-1)^2*1#

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

The slope of the tangent line is:

#f'(0)=-4/(0-1)^2=-4/(-1)^2=-4#

The equation of the line with a slope of #-4# passing through the point #(0,-4)#, which is the line's #y#-intercept, is

#y=-4x-4#

Graphed are #f(x)# and its tangent line at #x=0#:

graph{(y-4(x-1)^(-1))(y+4x+4)=0 [-15.73, 12.74, -11.85, 2.39]}