To find the range of the function $G(x) = \frac{x^2 + 5}{x + 1}$, let's proceed step by step:

Step 1: Analyze the function

The given function is a rational function of the form:

$G(x) = \frac{x^2 + 5}{x + 1}$

Here, the denominator $x + 1$ means that the function is undefined at $x = -1$, since division by zero is not allowed.

Step 2: Express $G(x)$ in a simpler form

Let's try dividing the numerator by the denominator using polynomial long division.

$\frac{x^2 + 5}{x + 1} = \text{(divide } x^2 \text{ by } x) = x$

Now, multiply $x$ by $x + 1$, which gives:

$x(x + 1) = x^2 + x$

Now subtract $(x^2 + x)$ from $(x^2 + 5)$:

$x^2 + 5 - (x^2 + x) = 5 - x$

So, we can rewrite $G(x)$ as:

$G(x) = x + \frac{5 - x}{x + 1}$

Step 3: Analyze asymptotic behavior

• As $x \to \infty$, $G(x) \approx x$ since the higher degree term in the numerator dominates the function.
• As $x \to -1$ from either side, the function approaches infinity or negative infinity, so there is a vertical asymptote at $x = -1$.

Step 4: Find the critical points

To find the critical points, calculate the derivative of $G(x)$:

$G'(x) = \frac{d}{dx} \left( \frac{x^2 + 5}{x + 1} \right)$

Using the quotient rule:

$G'(x) = \frac{(2x)(x+1) - (x^2 + 5)(1)}{(x + 1)^2}$

Simplifying the numerator:

$G'(x) = \frac{2x(x + 1) - (x^2 + 5)}{(x + 1)^2} = \frac{2x^2 + 2x - x^2 - 5}{(x + 1)^2} = \frac{x^2 + 2x - 5}{(x + 1)^2}$

Now, set $G'(x) = 0$ to find critical points:

$x^2 + 2x - 5 = 0$

Solve this quadratic equation:

$x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-5)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 20}}{2} = \frac{-2 \pm \sqrt{24}}{2} = \frac{-2 \pm 2\sqrt{6}}{2} = -1 \pm \sqrt{6}$

Thus, the critical points are:

$x = -1 + \sqrt{6}, \quad x = -1 - \sqrt{6}$

Step 5: Check the behavior at the critical points

At these critical points, you can evaluate the function to find the specific values at these points, which will help in determining the range.

Step 6: Conclusion

From the analysis, we can conclude that the range of $G(x)$ is all real numbers $y \in \mathbb{R}$ because the function's asymptotic behavior covers both positive and negative values as $x$ approaches large values and negative infinity.

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Would you like me to go further into the calculation of critical point values, or provide a detailed sketch? Here are a few related questions:

1. How do we find the domain of a rational function?
2. What happens at the vertical asymptote of this function?
3. How do critical points affect the graph of a function?
4. How is polynomial long division useful for simplifying rational functions?
5. What is the quotient rule for derivatives?

Tip: For rational functions, always check for vertical asymptotes and horizontal behavior to determine the range.