Posted by **Help asap!! plz** on Friday, October 22, 2010 at 6:11pm.

Consider the function:

f(x) = x/(x^2 + 2)

(a) Determine the intervals on which the function is concave up. Enter the leftmost interval first (e.g. (1,2) would come before (3,4)).

(b) Determine the intervals on which the function is concave down. Enter the leftmost interval first.

i only got to figure out the first derivative. And i did figure out second derivative but it came out to be some thing very long and nasty. so please help!

f'(x)=((x^2+2)-2x)/(x^2 + 2)^2

- CALCULUS -
**Reiny**, Friday, October 22, 2010 at 6:24pm
always simplify your first derivative before trying to take the second deriv.

I had it as

f' (x) = (2-x^2)/(x^2 + 2)^2

f'' (x) =[ (x^2+2)^2(-2x) - (2-x^2)(2)(x^2=2)(2x)]/ (x^2+2)^4

this factored to

-2x(x^2+2)[x^2+2 + 2(2-x^2)]/(x^2+2)^4

= -2x(6-x^2)/(x^2+2)^3

now the denominator cannot be zero and will always be positive, so as far as signs are concerned, we could ignore it.

so the curve is concave up when

-2x(6-x^2) > 0 or

2x(6-x^2) < 0

and the curve is concave down when

-2x(6-x^2) < 0 or

2x(6-x^2) > 0

can you take it from there?

btw, the points of inflection happen when

x = 0 or x = ±√6

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