At which point(s) does the graph of the function #f(x) = (x^2) / (x - 1)# have a horizontal tangent line?

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Answer 1 · 2016-10-20T01:21:18

#(0, 0), (2, 4)#

The slope of the tangent line of a graph #y=f(x)# at a point #x_0# is given by the derivative of #f# at that point, that is, #f'(x_0)#.

A horizontal tangent line implies a slope of #0#, so our goal is to find the points at which the derivative #f(x)# evaluates to #0#.

Using the quotient rule, we find the derivative as

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

#=((x-1)(d/dxx^2)-x^2(d/dx(x-1)))/(x-1)^2#

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

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

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

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

Setting this equal to zero, we get

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

#=> x(x-2) = 0#

#=> x = 0 or x = 2#

Thus, the graph of #f(x)# has a horizontal tangent line at #x=0# and #x=2#, that is, at the points

#(0, 0), (2, 4)#