What are the points of inflection of #f(x)=(2x-3)/(2x-1) #?

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Answer 1 · 2016-12-17T05:53:41

None.

By actual division,

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

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

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

#f'' does not vanish at all.

So, there is no point of inflexion.

As #(f-1)(x-1/2)=-1#, the graph is a rectangular hyperbole,

with asymptotes #x = 1/2 and y = 1#. Graph is inserted.

graph{(y-1)(x-1/2)+1=0x^2 [-10, 10, -5, 5]}