Question #148a8

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Answer 1 · 2017-11-13T16:41:17

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

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.

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