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# function

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If $y = \frac{x + 1}{x^2 + 1 - x},$ and $x$ is any real number, then what is the sum of the maximum and minimum possible values of $y \, ?$

Aug 29, 2023

#1
+177
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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!

Aug 30, 2023
#2
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4(2+2)=16

Guest Aug 31, 2023
#3
+170
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This problem does not require calculus to solve. We cross-multiply and simplify the given equation to get $$yx^2-(y+1)x+y-1=0$$. The domain of the function is all real numbers as $$x^2-x+1$$ does not have any real solutions, so the discriminant of the equation in $$x$$ must always be nonnegative, that is,
$$(y+1)^2-4(y)(y-1) \ge 0 \implies 3y^2-6y-1\le0.$$
The maximum and minimum are simply the roots of the LHS, and their sum is $$-\frac{-6}3=\boxed{-2}$$. The values of $$y$$ are achievable at the $$x$$-values found in the previous answer.

Aug 31, 2023
edited by plaintainmountain  Aug 31, 2023
edited by plaintainmountain  Aug 31, 2023
#4
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Oh, you mean that pathetic excuse for a solution that was ripped from ChatGPT and plastered all over the internet? Yeah, sure thing, it's obvious that you're too dumb to solve the problem on your own. I'm sure it must feel great relying on a stupid AI to give you all the answers for free. You should be ashamed of yourself, but then again, you probably don't even know what shame is. Just keep doing what you're doing and never bother asking anyone for help ever again.

GA

Guest Sep 2, 2023
edited by Guest  Sep 2, 2023
#5
+2487
-2

For the record: Post #4, signed by GA, is not by the real GA.

GA

--. .-

GingerAle  Sep 3, 2023
edited by GingerAle  Sep 3, 2023