1. ## Concavity

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?

2. Originally Posted by sw3etazngyrl
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?

3. I understand how to do the part you just showed me. But how do find out it's negative? I graphed it, and it's negative on the interval (0, infinity). Is that right? I know that [0, inifinity) is wrong.

4. Originally Posted by sw3etazngyrl
I understand how to do the part you just showed me. But how do find out it's negative? I graphed it, and it's negative on the interval (0, infinity). Is that right? I know that [0, inifinity) is wrong.
it is negative when $\displaystyle -(2x + 1) < 0$, that won't be the interval $\displaystyle (0, \infty)$

5. Given what you told about me about the inequality, would it be (2, infinity) since x>2?

6. sorry, i meant to put (1, infinity)

7. Sorry, again. I've been making careless mistakes. My final answer: (-1/2, infinity)

8. Originally Posted by sw3etazngyrl
Sorry, again. I've been making careless mistakes. My final answer: (-1/2, infinity)
yeah

9. Thank you so much!

10. ## the inflection point is at x=-1/2

Originally Posted by sw3etazngyrl
Sorry, again. I've been making careless mistakes. My final answer: (-1/2, infinity)
Yes, you are right. The inflection point is at x=-1/2 as shown by the figure below.

11. ## one more figure

one more figure

12. Originally Posted by curvature
one more figure
well, you cant really tell where the inflection point is on either graph.

13. Originally Posted by Jhevon
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?
Based on the calculation above:f"(x) changes signs at x=-1/2.
the figures just visualize the result.

14. Originally Posted by curvature
Based on the calculation above:f"(x) changes signs at x=-1/2.
the figures just visualize the result.
yeah, i know. i was just being nitpicky. i was in that kind of a mood. i'm bored to death right now. your figures definately helped. we may not be able to tell exactly what the inflection point is, but it does show that we could not be way off, which is good