Concavity

**sw3etazngyrl** · June 15th 2007, 03:48 PM

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?

**Jhevon** · June 15th 2007, 04:36 PM

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

$f(x) = \int_{0}^{x} \frac {1}{6 + t + t^2}dt$

$\Rightarrow f'(x) = \frac {d}{dt} \int_{0}^{x} \frac {1}{6 + t + t^2} dt$

$\Rightarrow f'(x) = \frac {1}{6 + x + x^2}$ by the second fundamental theorem of calculus

$\Rightarrow f''(x) = \frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2 }$

we have the curve being concave down when the second derivative is negative, so solve for that. that is,

on what interval is $\frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2}$ negative?

**sw3etazngyrl** · June 15th 2007, 04:38 PM

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.

**Jhevon** · June 15th 2007, 04:39 PM

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 $-(2x + 1) < 0$ , that won't be the interval $(0, \infty)$

**sw3etazngyrl** · June 15th 2007, 04:42 PM

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

**sw3etazngyrl** · June 15th 2007, 04:42 PM

sorry, i meant to put (1, infinity)

**sw3etazngyrl** · June 15th 2007, 04:44 PM

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

**Jhevon** · June 15th 2007, 04:44 PM

Originally Posted by sw3etazngyrl

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

yeah

**sw3etazngyrl** · June 15th 2007, 04:51 PM

Thank you so much!

**curvature** · June 15th 2007, 06:46 PM

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.

**curvature** · June 15th 2007, 07:19 PM

one more figure

**Jhevon** · June 15th 2007, 07:22 PM

Originally Posted by curvature

one more figure

well, you cant really tell where the inflection point is on either graph.

**curvature** · June 15th 2007, 07:28 PM

Originally Posted by Jhevon

We find concavity from the second derivative

$f(x) = \int_{0}^{x} \frac {1}{6 + t + t^2}dt$

$\Rightarrow f'(x) = \frac {d}{dt} \int_{0}^{x} \frac {1}{6 + t + t^2} dt$

$\Rightarrow f'(x) = \frac {1}{6 + x + x^2}$ by the second fundamental theorem of calculus

$\Rightarrow f''(x) = \frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2 }$

we have the curve being concave down when the second derivative is negative, so solve for that. that is,

on what interval is $\frac {-(2x + 1)}{ \left( 6 + x + x^2 \right)^2}$ negative?

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

**Jhevon** · June 15th 2007, 07:30 PM

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

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Concavity

the inflection point is at x=-1/2

one more figure

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