Find the interval on which the following curve is concave upward.?

Question

Find the interval on which the following curve is concave upward.

the integral of 1/ (1+ t+ t^2) over the interval (x,0)

Answer 1

if f(x) = ∫[t = x to 0] 1/ (1 + t + t²) dt

f'(x) = -1/ (1 + x + x²)

f''(x) = (1 + 2x)/(1 + x + x²)²

For concave UP f''(x) > 0

So 1 + 2x > 0 (as (1 + x + x²)² > 0 for all x)

ie x < -1/2

Answer 2

Interesting. Concavity is determined by looking at the second derivative. If you integrate and then differentiate twice, welll....

So, the first derivative of the integral is just the function with t replaced with x, right?

Differentiate again and find interval on which this is positive.


I'm a little suprized to see the interval as (x,0) rather than (0,x). This might have the effect of reversing the sign of the derivative.

In any case, don't do the integral. Write it out and differentiate taking care to handle the derivative of the limits. Then differentiate.

Answer 3

a) derive that and set it equal to 0 the solutions for x are the points b) do the same with the second derivative c) sketch some random ****