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

None.

By precise division,

#f'' does not vanish at all.

Therefore, there isn't a turning point.

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

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To find the points of inflection, we need to find the second derivative of the function and then determine where it changes sign. The second derivative of ( f(x) = \frac{{2x - 3}}{{2x - 1}} ) is ( f''(x) = \frac{{12}}{{(2x - 1)^3}} ). Setting the second derivative equal to zero, we find that it is undefined at ( x = \frac{1}{2} ). Therefore, the function has no points of inflection.

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