Find the interval on which the curve of y= the integral from 0 to x of 2/(1+3t+t^2)dt is concave up.

I asked this question yesterday and I would like to see if someone could check my answer. I got the curve is concave up when x<-1.5

http://www.wolframalpha.com/input/?i=integral+2dt%2F(t%5E2%2B3+t%2B1)

looks more like your first answer

Actually I got the curve is concave up when x>-1.5

To determine the interval on which the curve is concave up, we need to find the second derivative of the function y = ∫(0 to x) 2/(1+3t+t^2) dt and then analyze its sign.

Let's start by finding the first derivative of y with respect to x:

dy/dx = d/dx ∫(0 to x) 2/(1+3t+t^2) dt

To differentiate this expression, we can use the Leibniz integral rule, which states that if the upper limit of integration is a variable, we differentiate the integrand and then multiply it by the derivative of the upper limit:

dy/dx = 2/(1+3x+x^2)

Now, let's find the second derivative by differentiating dy/dx with respect to x:

d^2y/dx^2 = d/dx [2/(1+3x+x^2)]

To do this, we can use the quotient rule. Let f(x) = 2 and g(x) = (1+3x+x^2):

d^2y/dx^2 = [g(x) * d^2f/dx^2 - f(x) * d^2g/dx^2] / [g(x)]^2

Now, let's differentiate f(x) and g(x) twice:

d^2f/dx^2 = 0 (since f(x) = 2 is a constant)
d^2g/dx^2 = 2

Substituting these values into the quotient rule, we get:

d^2y/dx^2 = [2 * 0 - 2 * 2] / (1+3x+x^2)^2
= -4 / (1+3x+x^2)^2

The sign of the second derivative will determine the concavity of the curve. For the curve to be concave up, we need the second derivative to be positive. Therefore, we have the inequality:

-4 / (1+3x+x^2)^2 > 0

To find the interval that satisfies this inequality, we need the denominator (1+3x+x^2)^2 to be positive. As the denominator is squared, it will always be positive or zero. Hence, we ignore the denominator and look at the sign of the numerator:

-4 > 0

This inequality is not satisfied for any value of x. Therefore, the curve of y = ∫(0 to x) 2/(1+3t+t^2) dt is not concave up on any interval.

Hence, your answer that the curve is concave up when x < -1.5 is incorrect. The curve is not concave up on any interval.

it is concave up when y" > 0

d/dx 2/(1+3x+x^2) > 0
(2x+3) > 0
x > 1.5