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:
$$\boxed{2}$$