# Determining where the function is concave up? (Given an integral)

1. Apr 12, 2014

### JessicaJ283782

1. The problem statement, all variables and given/known data

x
Determine all intervals over which the function f(x)=∫√(1+t^2) dt is concave upward
1

2. Relevant equations

I know concave up means the f"(x)>0, so you have to get the second derivative

3. The attempt at a solution

f'(x)=√(1+x^) by susbituting in x

Then f"(x)=1/2*(1+x^2)^(-1/2)*2x

This is equal to:
2x/2√(1+x^2)

Then, I'm really confused on what to do next because the denominator is never zero, so would the only critical point be 0 and you check to the left and right of that?

2. Apr 12, 2014

### LCKurtz

Apparently that is $\int_1^x \frac 1 {\sqrt{1+t^2}}~dt$

Yes. Find where it is positive. Don't be afraid to cancel the 2's.

3. Apr 12, 2014

### JessicaJ283782

I apologize for the typo! So it would be:

x/√(1+x^2)

The only critical number would be 0 then, so wouldn't it be concave up on (-infinity, 0) and (0, infinity) since the denominator is always positive because of the square root?

4. Apr 12, 2014

### LCKurtz

Is the numerator always positive?

5. Apr 12, 2014

### JessicaJ283782

The numerator would give only a critical point of 0 since that is the only place it is not defined, right? So then it would be concave up on (0, infinity)?

6. Apr 12, 2014

### LCKurtz

There is no $x$ where that fraction is not defined. Instead of asking me if $(0,\infty)$ is the answer, explain why you think it is.

7. Apr 12, 2014

### JessicaJ283782

It would be (0, ∞) since the critical point is 0 because it makes the fraction zero. Concave up means f"(x)>0 and f"(x) is positive when the function is evaluated at any number greater than 0. If you evaluate the function at -1, for example, you would get a negative number, so it would be concave down less than 0. If that makes sense?

8. Apr 12, 2014

### LCKurtz

Yes, that makes perfect sense. Since $f''(x)=0$ when $x=0$, that is a point where the second derivative might change sign (or maybe not). That's why you check the signs on both sides. For $x<0$ you have $\frac - +$ and for $x>0$ you have $\frac + +$ for the signs. That's why $f''(x)>0$ for $x>0$. That's how you want to analyze problems like this.

9. Apr 12, 2014

### JessicaJ283782

Thank you so much! I really appreciate it!