Question

find the intervals on which f is increasing and decreasing
f(x)=(x-2)/(X^2-x+1)^2
I found the derivative a got a whole mess of things:
((x^2-x+1)^2-2(x^2-x+1)(2x-1)(x-2))/(x^2-x+1)^4

Do I have to use the quadratic formula to solve for the vertices to find where it is increasing and decreasing now?

Answer 1

You need to simplify your derivative a bit. It boils down to

-3(x^2-3x+1)/(x^2-x+1)^3

That's a bit easier to work with. See where that takes you.

Answer 2

I cannot figure out how to simplify this.

Answer 3

okay
(x^2-x+1)-2[(x^2-x+1)(2x-1)(x-2)]/(x^2-x+1)^4

(x^2-x+1)-2[(2x-1)(x-2)](x^2-x+1)/(x^2-x+1)^4

factor out (x^2-x+1)

(x^2-x+1)[-2(x-1)(x-2)]/(x^2-x+1)^4

-2(x^2-3x+2)/(x^2-x+1)^3

plz do wait for mathematician steve to come complete it for you
but that the simplification am sure

Answer 4

Once you have convinced yourself that the derivative really is

-3(x^2-3x+1)/(x^2-x+1)^3

f(x) is increasing where f'(x) is positive

Note that the denominator is always positive, so f' is positive when the numerator is positive. That is, when x^2-3x+1= is negative.

That is a parabola which opens upward, so it is negative between the roots. So, on the interval

((3-√5)/2,(3+√5)/2) or (0.38,2.62) is positive, meaning f is increasing, and decreasing elsewhere.

A look at the graph confirms this.

http://www.wolframalpha.com/input/?i=(x-2)%2F(x%5E2-x%2B1)%5E2

Note the local extrema, and f is increasing between the min and the max.