First off, I think you've differentiated it wrong. You've not used the chain rule properly when differentiating the bottom bracket. So let's start again.
![[Image: eq.latex?f(x)=\frac%7Bx-2%7D%7B(x%5E2-x+1)%5E2%7D]](http://codecogs.com/eq.latex?f(x)=\frac%7Bx-2%7D%7B(x%5E2-x+1)%5E2%7D)
Differentiating that by quotient and chain rule we get:
[IMG]http://codecogs.com/eq.latex?f'(x)&=\frac{(x^2-x+1)^2(1)-(x-2)(x^2-x+1)(2)(2x-1)}{(x^2-x+1)^4}[/IMG]
Which I simplify down to:
[IMG]http://codecogs.com/eq.latex?f'(x)=-\frac{3(x^2-3x+1)}{(x^2-x+1)^3}[/IMG]
We might as well get the 2nd derivative while we're here. Again, using quotient and chain rule:
[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{(x^2-x+1)^3(-3[2x-3])-(-3[x^2-3x+1])(x^2-x+1)^2(3)(2x-1)}{(x^2-x+1)^6}[/IMG]
Which simplifies to:
[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{6x(2x^2-8x+5)}{(x^2-x+1)^4}[/IMG]
Now you might want to double-check those derivatives, but I hope they're right lol.
Now for part (a), intervals for which it's an increasing function. As stated above, it's increasing when it's gradient is positive (i.e. f'(x) > 0), so:
[IMG]http://codecogs.com/eq.latex?f'(x)>0[/IMG]
![[Image: eq.latex?\Rightarrow%20-3(x%5E2-3x+1)%3E0]](http://codecogs.com/eq.latex?\Rightarrow%20-3(x%5E2-3x+1)>0)
![[Image: eq.latex?\Rightarrow%20x%5E2-3x+1%3C0]](http://codecogs.com/eq.latex?\Rightarrow%20x%5E2-3x+1<0)
Using the quadratic formula, I get (as you've previously mentioned):
![[Image: eq.latex?\frac%7B1%7D%7B2%7D(3\pm%20\sqrt%7B5%7D)]](http://codecogs.com/eq.latex?\frac%7B1%7D%7B2%7D(3\pm%20\sqrt%7B5%7D))
![[Image: eq.latex?\therefore\frac%7B1%7D%7B2%7D(3-\sqrt%7B5...+\sqrt%7B5%7D)]](http://codecogs.com/eq.latex?\therefore\frac%7B1%7D%7B2%7D(3-\sqrt%7B5%7D)<x<\frac%7B1%7D%7B2%7D(3+\sqrt%7B5%7D))
Right, now for part (b), intervals for which it's a decreasing function. Given that we've already done part (a), this is pretty trivial. It's either increasing, decreasing or stationary. Given that the two solutions above are the only stationary points (when f'(x) = 0), the opposite intervals must describe when the function is decreasing:
![[Image: eq.latex?\therefore%20x%3C\frac%7B1%7D%7B2%7D(3-\sqr...+\sqrt%7B5%7D)]](http://codecogs.com/eq.latex?\therefore%20x<\frac%7B1%7D%7B2%7D(3-\sqrt%7B5%7D)\And%20x>\frac%7B1%7D%7B2%7D(3+\sqrt%7B5%7D))
Part ©, intervals for which the function is concave up. As already said, when f''(x) > 0:
[IMG]http://codecogs.com/eq.latex?f''(x)>0[/IMG]
![[Image: eq.latex?\Rightarrow%206x(2x%5E2-8x+5)%3E0]](http://codecogs.com/eq.latex?\Rightarrow%206x(2x%5E2-8x+5)>0)
Either:
![[Image: eq.latex?6x=0\Rightarrow%20x=0]](http://codecogs.com/eq.latex?6x=0\Rightarrow%20x=0)
Or:
![[Image: eq.latex?2x%5E2-8x+5=0\Rightarrow%20x=2\pm%20\...2%7D\sqrt%7B6%7D]](http://codecogs.com/eq.latex?2x%5E2-8x+5=0\Rightarrow%20x=2\pm%20\frac%7B1%7D%7B2%7D\sqrt%7B6%7D)
Therefore:
![[Image: eq.latex?0%3Cx%3C2-\frac%7B1%7D%7B2%7D\sqrt%7B6%7D\And%20x...2%7D\sqrt%7B6%7D]](http://codecogs.com/eq.latex?0<x<2-\frac%7B1%7D%7B2%7D\sqrt%7B6%7D\And%20x>2+\frac%7B1%7D%7B2%7D\sqrt%7B6%7D)
Again, as with part (b), we can get part (d), intervals for which the function is concave down, from the above answer:
![[Image: eq.latex?\therefore%20x%3C0\And%202-\frac%7B1%7D%7B2...2%7D\sqrt%7B6%7D]](http://codecogs.com/eq.latex?\therefore%20x<0\And%202-\frac%7B1%7D%7B2%7D\sqrt%7B6%7D<x<2+\frac%7B1%7D%7B2%7D\sqrt%7B6%7D)
Finally, part (e), the co-ordinates of all inflection points. Again, this is trivial given our answers to parts © & (d), as we already know the 3 x-values for which f''(x) = 0:
![[Image: eq.latex?x=0\And%20x=2\pm%20\frac%7B1%7D%7B2%7D\sqrt%7B6%7D]](http://codecogs.com/eq.latex?x=0\And%20x=2\pm%20\frac%7B1%7D%7B2%7D\sqrt%7B6%7D)
Hope that's all correct, lol. If it's not, I at least hope that it helps.
Differentiating that by quotient and chain rule we get:
[IMG]http://codecogs.com/eq.latex?f'(x)&=\frac{(x^2-x+1)^2(1)-(x-2)(x^2-x+1)(2)(2x-1)}{(x^2-x+1)^4}[/IMG]
Which I simplify down to:
[IMG]http://codecogs.com/eq.latex?f'(x)=-\frac{3(x^2-3x+1)}{(x^2-x+1)^3}[/IMG]
We might as well get the 2nd derivative while we're here. Again, using quotient and chain rule:
[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{(x^2-x+1)^3(-3[2x-3])-(-3[x^2-3x+1])(x^2-x+1)^2(3)(2x-1)}{(x^2-x+1)^6}[/IMG]
Which simplifies to:
[IMG]http://codecogs.com/eq.latex?f''(x)=\frac{6x(2x^2-8x+5)}{(x^2-x+1)^4}[/IMG]
Now you might want to double-check those derivatives, but I hope they're right lol.
Now for part (a), intervals for which it's an increasing function. As stated above, it's increasing when it's gradient is positive (i.e. f'(x) > 0), so:
[IMG]http://codecogs.com/eq.latex?f'(x)>0[/IMG]
Using the quadratic formula, I get (as you've previously mentioned):
Right, now for part (b), intervals for which it's a decreasing function. Given that we've already done part (a), this is pretty trivial. It's either increasing, decreasing or stationary. Given that the two solutions above are the only stationary points (when f'(x) = 0), the opposite intervals must describe when the function is decreasing:
Part ©, intervals for which the function is concave up. As already said, when f''(x) > 0:
[IMG]http://codecogs.com/eq.latex?f''(x)>0[/IMG]
Either:
Or:
Therefore:
Again, as with part (b), we can get part (d), intervals for which the function is concave down, from the above answer:
Finally, part (e), the co-ordinates of all inflection points. Again, this is trivial given our answers to parts © & (d), as we already know the 3 x-values for which f''(x) = 0:
Hope that's all correct, lol. If it's not, I at least hope that it helps.