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

Jun 18, 2024

#2
+129820
+1

y =  (x + 1) (x^2 + 1)^-1

y' = (x^2 + 1)^-1  - (x +1) (x^2 + 1)^-2 * 2x  = 0

(x^2 + 1)^-2  [  (x^2 + 1) - 2x ( x  + 1) ]  = 0

-x^2  - 2x + 1   = 0

x^2 + 2x - 1  =  0

x^2 + 2x  = 1

x^2 + 2x + 1  =  2

(x + 1)^2   =  2       take both roots

x + 1 = sqrt 2            x + 1  = -sqrt 2

x = sqrt (2)  - 1         x = -sqrt (2)  - 1

y  =  sqrt 2 / [ (sqrt (2) -1)^2 + 1]  = sqrt 2 / [ 4 - 2sqrt 2]  =  max

y = -sqrt (2) / [ (-sqrt (2) - 1)^2 + 1 ]  = -sqrt (2) / [ 4 + 2sqrt 2] = min

Sum of  max and  min  = [ sqrt 2][ 4 + sqrt 8] / 8  - [sqrt 2][ 4 -sqrt 8] /8  =

2sqrt (16) / 8   =

[2 * 4] / 8  =

8  / 8    =

1

Jun 19, 2024
edited by CPhill  Jun 19, 2024

#1
+1459
0

To determine the maximum and minimum values of the function $$y = \frac{x+1}{x^2+1}$$, we first find the critical points by differentiating $$y$$ with respect to $$x$$.

Given:

$y = \frac{x + 1}{x^2 + 1}$

Using the quotient rule for differentiation:

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

Calculate the derivatives:
$\frac{d}{dx}(x + 1) = 1$

$\frac{d}{dx}(x^2 + 1) = 2x$

Substitute these into the quotient rule:

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

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

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

Set the derivative $$y'$$ to zero to find the critical points:

$\frac{-x^2 - 2x + 1}{(x^2 + 1)^2} = 0$

The numerator must be zero:

$-x^2 - 2x + 1 = 0$

$x^2 + 2x - 1 = 0$

Using the quadratic formula $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ with $$a = 1$$, $$b = 2$$, and $$c = -1$$:
$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}$

$x = \frac{-2 \pm \sqrt{4 + 4}}{2}$

$x = \frac{-2 \pm \sqrt{8}}{2}$

$x = \frac{-2 \pm 2\sqrt{2}}{2}$

$x = -1 \pm \sqrt{2}$

Thus, the critical points are:
$x = -1 + \sqrt{2} \quad \text{and} \quad x = -1 - \sqrt{2}$

Evaluate $$y$$ at these critical points:

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

$= \frac{\sqrt{2}}{(1 - 2\sqrt{2} + 2 + 1)}$

$= \frac{\sqrt{2}}{4 - 2\sqrt{2}}$

Rationalize the denominator:

$= \frac{\sqrt{2}(4 + 2\sqrt{2})}{(4 - 2\sqrt{2})(4 + 2\sqrt{2})}$

$= \frac{4\sqrt{2} + 4\cdot 2}{16 - 8}$

$= \frac{4\sqrt{2} + 8}{8}$

$= \frac{4\sqrt{2}}{8} + 1$

$= \frac{\sqrt{2}}{2} + 1$

Next, evaluate $$y$$ at $$x = -1 - \sqrt{2}$$:

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

$= \frac{-\sqrt{2}}{(1 + 2\sqrt{2} + 2 + 1)}$

$= \frac{-\sqrt{2}}{4 + 2\sqrt{2}}$

Rationalize the denominator:

$= \frac{-\sqrt{2}(4 - 2\sqrt{2})}{(4 + 2\sqrt{2})(4 - 2\sqrt{2})}$

$= \frac{-4\sqrt{2} + 4 \cdot 2}{16 - 8}$

$= \frac{-4\sqrt{2} + 8}{8}$

$= \frac{-4\sqrt{2}}{8} + 1$

$= -\frac{\sqrt{2}}{2} + 1$

So, the maximum value is $$\frac{\sqrt{2}}{2} + 1$$ and the minimum value is $$-\frac{\sqrt{2}}{2} + 1$$. The sum of the maximum and minimum values is:

$\left( \frac{\sqrt{2}}{2} + 1 \right) + \left( -\frac{\sqrt{2}}{2} + 1 \right) = 1 + 1 = 2$

Thus, the sum of the maximum and minimum possible values of $$y$$ is:
$\boxed{2}$

Jun 18, 2024
#2
+129820
+1

y =  (x + 1) (x^2 + 1)^-1

y' = (x^2 + 1)^-1  - (x +1) (x^2 + 1)^-2 * 2x  = 0

(x^2 + 1)^-2  [  (x^2 + 1) - 2x ( x  + 1) ]  = 0

-x^2  - 2x + 1   = 0

x^2 + 2x - 1  =  0

x^2 + 2x  = 1

x^2 + 2x + 1  =  2

(x + 1)^2   =  2       take both roots

x + 1 = sqrt 2            x + 1  = -sqrt 2

x = sqrt (2)  - 1         x = -sqrt (2)  - 1

y  =  sqrt 2 / [ (sqrt (2) -1)^2 + 1]  = sqrt 2 / [ 4 - 2sqrt 2]  =  max

y = -sqrt (2) / [ (-sqrt (2) - 1)^2 + 1 ]  = -sqrt (2) / [ 4 + 2sqrt 2] = min

Sum of  max and  min  = [ sqrt 2][ 4 + sqrt 8] / 8  - [sqrt 2][ 4 -sqrt 8] /8  =

2sqrt (16) / 8   =

[2 * 4] / 8  =

8  / 8    =

1

CPhill Jun 19, 2024
edited by CPhill  Jun 19, 2024
#3
+85
+1

Thanks so much!

PurpleWasp  Jun 19, 2024