Application of second derivative S(t)= 3t / (t^2 +2)

[cmnalo]

I need to find where the function is concave upward and where it is concave downward.

S(t)= 3t / (t^2 +2)

S'(t) = -3 / (t^2 + 2)
S"(t) = 6t / (t^2 + 2)^2

I know I need to set up a sign chart with the intervals but I'm lost and don't now how to get the intervals from this derivative.

The answer is (-squareroot of 6 , 0) U (square root of 6 , infinity)[/code]

 

[tkhunny]

You may wish to take another crack at those derivatives.

S(t) = 3t / (t^2 +2)

\(\displaystyle S'(t) = \frac{[(t^{2} +2)*3 - 3t*2t]}{(t^{2} +2)^{2}} = \frac{3(2-t^{2})}{(t^{2}+2)^{2}}\)

Okay, let's see if you can get a little closer on the second derivative.

 

[skeeter]

to begin with, you have to correctly find the first derivative ...

s(t) = 3t/(t^2 + 2)

quotient rule, correct? ...

s'(t) = [(t^2 + 2)3 - 3t(2t)]/(t^2 + 2)^2

s'(t) = (3t^2 + 6 - 6t^2)/(t^2 + 2)^2

s'(t) = (6 - 3t^2)/(t^2 + 2)^2

s'(t) = 3(2 - t^2)/(t^2 + 2)^2

now ... use the quotient rule again (correctly this time), you should get

s"(t) = 6t(t^2 - 6)/(t^2 + 2)^3 edit: fixed my sign

 

[cmnalo]

Wouldn't I also need to use the general power rule when appling the quotient rule since the denomonator is squared?

s'(t) = 3(2 - t^2)/(t^2 + 2)^2


s"(t) = (t^2 + 2)^2 d/dx 3(2 - t^2) - 3(2 - t^2) d/dx (t^2 + 2)^2

s"(t) = (t^2 + 2)^2 (2t) - 3(2 - t^2) [2(t^2 +2)(2t)] / [(t^2 + 2)^2]^2

S" (t) = (2t) - 3(2 - t^2) [2(t^2 +2)(2t)] / [(t^2 + 2)^2]

s" (t) = ?
I must be appling this rule incorrectly since this problem is making me dizzy. Please help.

 

[skeeter]

s'(t) = 3(2 - t^2)/(t^2 + 2)^2

s"(t) = [(t^2 + 2)^2*(-6t) - 3(2 - t^2)*2(t^2 + 2)*(2t)]/(t^2 + 2)^4

s"(t) = [-6t(t^2 + 2)^2 - 12t(2 - t^2)(t^2 + 2)]/(t^2 + 2)^4

factor out (t^2 + 2) from each term in the numerator ...

s"(t) = (t^2 + 2)[-6t(t^2 + 2) - 12t(2 - t^2)]/(t^2 + 2)^4

s"(t) = (-6t^3 - 12t - 24t + 12t^3)/(t^2 + 2)^3

s"(t) = (6t^3 - 36t)/(t^2 + 2)^3 = 6t(t^2 - 6)/(t^2 + 2)^3