Let f(x)= 1/(5x^2+4). Find the open intervals on which f is concave up (down). Then determine the x-coordinate

let f(x)= 1/(5x^2+4). Find the open intervals on which f is concave up (down). Then determine the x-coordinates of all inflection points of f.

1.
f is concave up on the intervals

2.
f is concave down on the intervals

3.
The inflection points occur at x =

y = (5x² + 4)^(-1)
y' = -(5x² + 4)^(-2) * 10x

y'' = 2(5x² + 4)^(-3) * 10x * 10x - (5x² + 4)^(-2) * 10
y'' = [2(10x)² - 10(5x² + 4)]/(5x² + 4)³
y'' = (200x² - 50x² - 40)/(5x² + 4)³
y'' = (150x² - 40)/(5x² + 4)³
y'' = 10(15x² - 4)/(5x² + 4)³

EDIT: You can also use the quotient rule twice to get this derivative, if you want.

So the inflection point happens at...

15x² - 4 = 0
x² = 4/15
x = ± 2/√15

That's the location of your inflection points. So now test values around them to find concavity. Test something to the left of -2/√15, something between -2/√15 and 2/√15, and something to the right of 2/√15. Try -1, 0, and 1.

At x = -1...

y''(-1) = 10(15 - 4)/(5 + 4)³ = (10)(11)/9³: positive, concave up
y''(0) = 10(-4)/(4)³: negative, concave down
y''(1) = 10(15 - 4)/(5 + 4)³ = (10)(11)/9³: positive, concave up

So the function is concave up from -∞ to -2/√15, concave down from -2/√15 to 2/√15, and concave up again from 2/√15 to +∞.

Graph: http://mathway.com/answer.aspx?p=grap?p=…

Inflection Point Check: http://www.wolframalpha.com/input/?i=inf…