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

There are 3 points of inflexion

diagram{x/(1+x^2) [-5, 5, -2.5, 2.5]}

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To find the points of inflection of ( f(x) = \frac{x}{1 + x^2} ), we need to locate where the concavity changes.

- Find the second derivative of ( f(x) ).
- Set the second derivative equal to zero and solve for ( x ) to find potential points of inflection.
- Test the concavity of ( f(x) ) around each potential point of inflection to confirm whether it changes.

First, find the first derivative:

[ f(x) = \frac{x}{1 + x^2} ] [ f'(x) = \frac{(1 + x^2)(1) - (x)(2x)}{(1 + x^2)^2} ] [ f'(x) = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} ] [ f'(x) = \frac{1 - x^2}{(1 + x^2)^2} ]

Now, find the second derivative:

[ f''(x) = \frac{(1 + x^2)^2(0) - (1 - x^2)(2(1 + x^2)(2x))}{(1 + x^2)^4} ] [ f''(x) = \frac{-4x(1 - x^2)}{(1 + x^2)^3} ]

Set ( f''(x) = 0 ) to find potential points of inflection:

[ -4x(1 - x^2) = 0 ]

Solve for ( x ):

[ -4x(1 - x^2) = 0 ] [ -4x + 4x^3 = 0 ] [ 4x^3 - 4x = 0 ] [ 4x(x^2 - 1) = 0 ] [ x(x - 1)(x + 1) = 0 ]

So, potential points of inflection are ( x = 0, x = 1, x = -1 ).

Test the concavity around each potential point of inflection to confirm if they are points of inflection. If the concavity changes signs around a point, it is a point of inflection.

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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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