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

Dec 17, 2016

None.

#### Explanation:

By actual division,

$f = 1 - \frac{1}{x - \frac{1}{2}}$

$f ' = \frac{1}{x - \frac{1}{2}} ^ 2$

$f ' ' = - \frac{2}{x - \frac{1}{2}} ^ 3$

#f'' does not vanish at all.

So, there is no point of inflexion.

As $\left(f - 1\right) \left(x - \frac{1}{2}\right) = - 1$, the graph is a rectangular hyperbole,

with asymptotes $x = \frac{1}{2} \mathmr{and} y = 1$. Graph is inserted.

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