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}
Am needing any help... Am having trouble finding the 2nd derivative
Here's what ive graphed
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)