# Thread: Urgent help on concavity

1. ## Urgent help on concavity

Find any points of inflection of the graph of the function and determine whether the graph is concave upward or concave downward:
G(x) = \frac{x}{x^2+4}
Am needing any help... Am having trouble finding the 2nd derivative
Here's what ive graphed

Find any points of inflection of the graph of the function and determine whether the graph is concave upward or concave downward:
G(x) = \frac{x}{x^2+4}
Use the quotient rule,
G'(x)=\frac{(x)'(x^2+4)-(x)(x^2+4)'}{(x^2+4)^2}
Thus,
G'(x)=\frac{x^2+4-x(2x)}{(x^2+4)^2}
Thus,
G'(x)=\frac{-x^2+4}{(x^2+4)^2}
Now the second derivative,
G''(x)=\frac{(-x^2+4)'(x^2+4)^2-(-x^2+4)[(x^2+4)^2]'}{(x^2+4)^4}
Thus,
G''(x)=\frac{-2x(x^2+4)^2-4x(-x^2+4)(x^2+4)}{(x^2+4)^4}
Factor,
G''(x)=\frac{-2x(x^2+4)[(x^2+4)+2(-x^2+4)}{(x^2+4)^4}
Thus,
G''(x)=\frac{-2x[x^2+4-2x^2+8]}{(x^2+4)^4}
Thus,
G''(x)=\frac{-2x(-x^2+12)}{(x^2+4)^3}

The inflection points is when G''(x)=0
That is when the numerator is zero.
Thus,
-2x(-x^2+12)=0
x=0
x=+/-sqrt(12)