Question

Question #148a8

Answer 1

`x=0` and `x=+-sqrt(3)~~+-1.732`

Explanation:

The inflection point of a function `f(x)` is where the second derivative is equal to zero.

We are given `f(x)=x/(x^2+1)`

The first derivative is determined using the quotient rule.

`f'(x)=((x^2+1)d/dx(x)-x d/dx(x^2+1))/((x^2+1)^2)`

`" "=(x^2+1-x(2x))/((x^2+1)^2)`

`" "=(1-x^2)/((1+x^2)^2)`

Next, we need to find the second derivative, again using the quotient rule.

`f^('')(x)=((1+x^2)^2 d/dx(1-x^2)-(1-x^2)d/dx((1+x^2)^2))/((1+x^2)^4)`

`" "=((1+x^2)^2 (-2x)-(1-x^2)2(1+x^2)2x)/((1+x^2)^4)`

`" "=(2x(1+x^2)(-(1+x^2)-2(1-x^2)))/((1+x^2)^4)`

`" "=(2x(1+x^2)(-1-x^2-2+2x^2))/((1+x^2)^3)`

`" "=(2x(-3+x^2))/((1+x^2)^2)`

The inflection point is where this second derivative is equal to zero. This last quotient is equal to zero whenever the numerator is equal to zero.

`2x(-3+x^2)=0`

`2x=0` and `-3+x^2=0`

`x=0` and `x=+-sqrt(3)~~+-1.732`

Finally, you can see this is where the original function changes it's concavity at those three points.