+189 We can find the maximum and minimum of \(y = \frac{x + 1}{x^2 - x + 1}\) by taking the derivative and determining what x-values make the derivative zero. To be honest, I always have trouble remembering the derivative quotient rule. I have to rehearse "Low D-High minus High D-Low" so that I can recall the derivative quotient rule every time!
\(y = \frac{x + 1}{x^2 - x + 1} \\ \frac{\text{d}y}{\text{d}x} = \frac{(x^2 - x + 1)(1) - (x + 1)(2x - 1)}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{x^2 - x + 1 - (2x^2 - x + 2x - 1)}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{x^2 - x + 1 - 2x^2 -x + 1}{(x^2 - x + 1)^2} \\ \frac{\text{d}y}{\text{d}x} = \frac{-x^2 -2x + 2}{(x^2 - x + 1)^2}\)
Now that we have the derivative, we want to identify the critical points, which are the x-values of possible maxima and minima of this particular equation. This can be accomplished by setting the derivative to zero and solving for x.
\(\frac{-x^2 - 2x + 2}{(x^2 -x + 1)^2} = 0 \\ -x^2 -2x + 2 = 0 \\ x^2 + 2x = 2 \\ x^2 + 2x + 1 = 2 + 1 \\ (x + 1)^2 = 3 \\ |x + 1| = \sqrt{3} \\ x_1 = \sqrt{3} - 1 \text{ or } x_2 = -\sqrt{3} - 1\)
We have successfully identified that these x-values are potentials for extrema, but we still have to confirm that these x-values indeed are extrema of this rational function. To do this, we will identify the sign of the first derivative of the x-values to the left and right of the critical points. If the sign changes, then we have identified a maximum or minimum. Since we have found the zeroes of the rational function above, we can factor the numerator. The denominator is a squared quantity, so the quantity will always be positive over the real numbers.
\(\frac{(x-(-\sqrt{3} - 1)(x - (\sqrt{3} - 1))}{(x^2 -x + 1)^2}\)
| Left | Zero | Right | Extrema? |
| + | \(x_1 = \sqrt{3} - 1\) | - | Since the sign changes from postiive to negative, this extremum is a maximum. |
| - | \(x_2 = -\sqrt{3} - 1\) | + | Since the sign changes from negative to positive, this extremum is a minimum. |
Now that we have confirmed that these critical points are indeed the maximum and minimum of this equation, we can substitute these values into the original equation and find the difference of the y-values.
| \(\text{Let } x = \sqrt{3} - 1: \\ y = \frac{\sqrt{3} - 1 + 1}{(\sqrt{3} - 1)^2 - (\sqrt{3} - 1) + 1} \\ y = \frac{\sqrt{3}}{3 - 2\sqrt{3} + 1 - \sqrt{3} + 1 + 1} \\ y = \frac{\sqrt{3}}{-3\sqrt{3}+6}\) | \(\text{Let } x = -\sqrt{3} - 1: \\ y = \frac{-\sqrt{3} - 1 + 1}{(-\sqrt{3} - 1)^2 - (-\sqrt{3} - 1) + 1} \\ y = \frac{-\sqrt{3}}{3+2\sqrt{3} + 1 + \sqrt{3} + 1 + 1} \\ y = \frac{-\sqrt{3}}{3\sqrt{3}+6}\) |
The question asks for the sum of the maximum and the minimum points.
\(\begin{align*} \frac{\sqrt{3}}{6 - 3\sqrt{3}} - \frac{\sqrt{3}}{6 + 3\sqrt{3}} &= \frac{\sqrt{3}}{6 - 3\sqrt{3}} * \frac{6 + 3\sqrt{3}}{6 + 3\sqrt{3}} - \frac{\sqrt{3}}{6 + 3\sqrt{3}} * \frac{6 - 3\sqrt{3}}{6 - 3\sqrt{3}} \\ &= \frac{6\sqrt{3} + 3 * 3}{36 - 9 * 3} - \frac{6\sqrt{3} - 3 * 3}{36 - 9 * 3} \\ &= \frac{6\sqrt{3} - 6 \sqrt{3} + 9 + 9}{9} \\ &= \frac{18}{9} \\ &= 2 \end{align*}\)
I am nearly in disbelief that the sum simplifies so nicely!
