Question

@perl @Jhannybean please help!

Answer 1

\[\frac{d}{dx}\int_0^x \frac{6}{1+2t+t^2}dt = \frac{6}{1+2x+x^2}=f(x)\]

Answer 2

yeah that would be the first derivative right?

Answer 3

No taking the derivative of the integral at the limits just gives us the function at the limits itself, Fundamental Theorem of Calculus.

Answer 4

ohhhh

Answer 5

So the first derivative becomes \[f(x) = 6(1+2x+x^2)^{-1}\]\[f'(x) = -\frac{12}{(x+1)^3}\]That is the first derivative, not the second derivative.

Answer 6

ohh!

Answer 7

\[f''(x) = \frac{36}{(x+1)^4}\]

Answer 8

got it(:

Answer 9

And now to find the interval where it is concave up (because \((x+1)^{\color{red}4}\)) you just find where \[(x+1)^4 = 0\]\[x+1 =0 \implies x = -1\]

Answer 10

so the interval is -1?

Answer 11

thank you!!!

Answer 12

\[(-\infty,-1)\sf \text{U}(-1,+\infty)\]