I need help on My Calc HW

Stuff like this is why people hate math. You've done your first derivative wrong - make sure you're doing the chain rule correctly.

 Spoiler

Use quotient rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(1)(x^2-x+1)^2-\frac{d}{dx}(x^2-x+1)^2(x-2)}{(x^2-x+1)^4}[/img]

Use chain rule.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)^2-2(x^2-x+1)(2x-1)(x-2)}{(x^2-x+1)^4}[/img]

Simplify.

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x-1)(x-2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{(x^2-x+1)-2(2x^2-5x+2)}{(x^2-x+1)^3}[/img]

[img]http://www.codecogs.com/eq.latex?f'(x) = \frac{-3(x^2-3x+1)}{(x^2-x+1)^3}[/img]

At the risk of bad algebra, this should be the first derivative.

Also, your Leibnitz notation is wrong. It's , not .

What are you asking for exactly? The answer itself? I'm not convinced you know what "inflection point", "concave up", and "concave down" even mean.

First: Thanks

Second: Well the book said:
Find: (a) the intervals on which F is increasin, (b) the intervals on which F is decresing, © the open intervals on which F is concave up, (d) the open intervals on which F is concave down, and (e) the x-coodinates of all inflection points.

Third i need both the simplified form of the first and 2nd derivative set each = to 0 and solve.. i did that to the one u gave me and I used to quad formula and got

3(+ or -)(square root 5)
2

now the answers for this problem is in the back and that turned (the fraction above) is the answer for A but i'm not 100% sure WHY it's the answer or why b,c,and d are the answers they are... i know E is the solved version of the 2nd derivative (solved that way shown above)