Please Help, Quadractics

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

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:

\[
-x^2 - 2x + 1 = 0
\]

Solve the quadratic equation:
\[
x^2 + 2x - 1 = 0
\]

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:
\[
\boxed{2}
\]