1. ## Find x-coordinates

Find the x-coordinates of the two stationary points on the graph of y = (2x-1)/(x^2+1).

How is this question solved. As far as I can get is using the quotient rule to differentiate y to give:
= 2(x^2+1)-2x(2x-1) All over (x^2+1)^2

How do I find the stationary points from this?

2. Originally Posted by haku
Find the x-coordinates of the two stationary points on the graph of y = (2x-1)/(x^2+1).

How is this question solved. As far as I can get is using the quotient rule to differentiate y to give:
= 2(x^2+1)-2x(2x-1) All over (x^2+1)^2

How do I find the stationary points from this?
$y = \frac{2x - 1}{x^2 + 1}$

$y^{\prime} = \frac{2 \cdot (x^2 + 1) - (2x - 1)(2x)}{(x^2 + 1)^2}$

$y^{\prime} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2 + 1)^2}$

$y^{\prime} = \frac{2(-x^2 + x + 1)}{(x^2 + 1)^2}$

Stationary points are where the derivative is equal to 0. So set this equal to 0, then solve.

-Dan

3. Would you use the quandratic formual to solve this? If so how would you find the values of a, b and c from the equation:

2(-x^2+x+1)=(x^2+1)^2

Would you apply the formula to (-x^2+x+1) ?

4. Originally Posted by haku
Would you use the quandratic formual to solve this? If so how would you find the values of a, b and c from the equation:

2(-x^2+x+1)=(x^2+1)^2
It's simpler than that. You made a slight error in your equation:
$\frac{2(-x^2 + x + 1)}{(x^2 + 1)^2} = 0$

Multiply both sides by $(x^2 + 1)^2$:
$2(-x^2 + x + 1) = 0$

Can you take it from here?

-Dan

5. I get 1 + (root5)/2 Or 1 - (root5)/2

Is this correct?

6. Originally Posted by haku
I get 1 + (root5)/2 Or 1 - (root5)/2

Is this correct?
If this is meant to be
$\frac{1 \pm \sqrt{5}}{2}$
then, yes, you have it right. (Please use parenthesis!)

-Dan