# Find interval

1. Nov 30, 2005

### powp

Hello
How do I do the following
Find the interval on which the curve
$$y = \int_x^0 \frac{1}{1 + t + t^2} dt$$
is concave upward.

any help would be great.

P

Just in case Latex does not show up
x
Large S 1/1(1 + t + t^2) dt
0

2. Nov 30, 2005

### Galileo

When, in general, is some function f concave upwards?

3. Nov 30, 2005

### powp

I belive a function f should be concave upward when f''(c) > 0. Am I correct. Don't have my text book on me.

4. Nov 30, 2005

### BobG

Correct.

Now, if the function f (t) is $$y = \int_x^0 \frac{1}{1 + t + t^2} dt$$, then what's f'(t). (This is very easy).

Finding f''(t) is a little harder. Once found set it up as an inequality and solve.

5. Nov 30, 2005

### powp

f' is (x)$$y = \int_x^0 \frac{-1-2t}{(1 + t + t^2)^2}$$

f'' would then be
$$y = \int_x^0 \frac{-1 + 2t + 2t^2}{(1 + t + t^2)^4} dt$$

is this correct??

How do I solve an inequality that is this complex(it is complex to me)?? I am really not sure about this.

6. Nov 30, 2005

### BobG

The integral is also called the anti-derivative, so the derivative of $$y = \int_x^0 \frac{1}{1 + t + t^2} dt$$
is just:
$$f'(t)= \frac{1}{1 + t + t^2}$$

So your second derivative is just:
$$f''(t) =\frac{-1-2t}{(1 + t + t^2)^2}$$

7. Nov 30, 2005

### Zurtex

Not quite BobG, think about where the x is and what your actually differentiating.

8. Nov 30, 2005

### powp

anybody with some guidence?

9. Nov 30, 2005

### Zurtex

Erm, you've been given loads and loads, what have you done with what you have been given?

10. Nov 30, 2005

### HallsofIvy

Staff Emeritus
"Fundamental Theorem of Calculus"!!!

What is the derivative of $$\int_a^x f(t)dt$$ according to the Fundamenta Theorem of Calculus? (This is what Bobg was doing.)

Knowing that, what is the derivative of $$\int_x^a f(t)dt$$? (This is what Bobg should have done!)

What is the derivative of $$\int_x^a f(t)dt$$