What are the #x#-coordinate(s) of the point(s) of inflection of #f(x) = x/(x^2+1)#?

1 Answer
May 30, 2017

#x={-sqrt(3),0,sqrt(3)}#

Explanation:

A point of inflection is the point at which a graph switches concavity or, more technically, where the second derivative (which describes concavity) is 0. Therefore we must first take the second derivative.

So, to take the first derivative use the quotient rule. This means the derivative of two functions which form a fraction is the bottom function multiplied by the top function's derivative minus the top function multiplied by the bottom fraction's derivative all over the bottom function squared. This gives us: #((x^2+1)-(x)*(2x))/(x^2+1)^2#. Or more simply: #(1-x^2)/(x^2+1)^2#.

Now we must take the derivative of THIS using the same technique. This gives us: #(((x^2+1)^2)(-2x)-(1-x^2)(2(x^2+1)*2x))/((x^2+1)^2)^2#. Or more simply: #(2x(x^2+1)(x^2-3))/(x^2+1)^4#.

Finally, we set this second derivative equal to 0 in order to find the x-coordinated of the points of inflection. This will be whenever/wherever the numerator is 0 (as long as the denominator is not 0 at the same time). This gives us the equation: #2x(x^2+1)(x^2-3)=0#. Using the product property we set every quantity equal to zero to get three separate equations: #2x=0#, #x^2+1=0#, and #x^2-3=0#. Lastly we solve all of the equations to get #x=0#, #x=+-sqrt(-1)#, and #x=+-sqrt(3)#. One of these is imaginary, so we throw it out because we are looking for real answers.

This gives us a final solution of #x={-sqrt(3),0,sqrt(3)}#.