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

Answer 1

None.

By precise 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.

Therefore, there isn't a turning point.

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

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Answer 2

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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