Find the interval on which the curve f(x)=integral of 1/(6+t+t^2)dt from 0 to x is concave downward.
Is the answer: (0, infinity)? If not, can you show me?
We find concavity from the second derivative
$\displaystyle f(x) = \int_{0}^{x} \frac {1}{6 + t + t^2}dt$
$\displaystyle \Rightarrow f'(x) = \frac {d}{dt} \int_{0}^{x} \frac {1}{6 + t + t^2} dt$
$\displaystyle \Rightarrow f'(x) = \frac {1}{6 + x + x^2}$ by the second fundamental theorem of calculus
$\displaystyle \Rightarrow f''(x) = \frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2 }$
we have the curve being concave down when the second derivative is negative, so solve for that. that is,
on what interval is $\displaystyle \frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2}$ negative?