range of G(×)=×²+5/×+1 {*x*∈R∣*x*=?}

We are given the function:

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

### Step 1: Domain of the Function
We need to determine when \( G(x) \) is undefined. Since division by zero is not allowed, we find the value of \( x \) that makes the denominator zero:

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]

Thus, the function is undefined at \( x = -1 \), so the domain of \( G(x) \) is all real numbers except \( x = -1 \), i.e., \( x \in \mathbb{R} \setminus \{-1\} \).

### Step 2: Behavior at \( x = -1 \)
To understand the range of the function, we will study the behavior of \( G(x) \) near \( x = -1 \). As \( x \) approaches \( -1 \) from the left (\( x \to -1^- \)) and the right (\( x \to -1^+ \)), the function tends to infinity or negative infinity, suggesting a vertical asymptote at \( x = -1 \).

### Step 3: Large \( x \) Behavior (Asymptotic Analysis)
For very large positive or negative values of \( x \), the quadratic term \( x^2 \) dominates both the numerator and the denominator. Therefore, the function simplifies asymptotically as follows:

\[
G(x) \approx \frac{x^2}{x} = x
\]

Thus, as \( x \to \infty \), \( G(x) \to \infty \), and as \( x \to -\infty \), \( G(x) \to -\infty \).

### Step 4: Critical Points
To find any minimum or maximum values of \( G(x) \), we calculate its derivative using the quotient rule:

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

Simplifying the numerator:

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

To find critical points, we set \( G'(x) = 0 \):

\[
x^2 + 2x - 5 = 0
\]

Solving this quadratic equation using the quadratic formula:

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

So, the critical points are \( x = -1 + \sqrt{6} \) and \( x = -1 - \sqrt{6} \).

### Step 5: Range of the Function
By analyzing the behavior at large \( x \) and the critical points, we observe that the function takes all real values except there may be some discontinuities. Due to the vertical asymptote at \( x = -1 \), \( G(x) \) tends towards infinity and negative infinity, so the range of the function is:

\[
\text{Range}(G) = \mathbb{R}
\]

### Conclusion:
The domain of the function is \( x \in \mathbb{R} \setminus \{-1\} \), and the range is all real numbers, i.e., \( \mathbb{R} \).

Would you like further clarification on any of the steps?

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### Related Questions:
1. What are the steps to solve rational functions in general?
2. How can we find horizontal asymptotes for rational functions?
3. What does the derivative tell us about the behavior of a function?
4. How do critical points relate to the maximum and minimum values of a function?
5. How can the quadratic formula be applied to different types of equations?

**Tip:** When dealing with rational functions, always check the domain by identifying where the denominator equals zero.

range of G(x)=x²+5/(x+1) {x∈R | x≠?}

Explore how to find the range of the rational function G(x) = (x² + 5)/(x + 1). This problem covers essential calculus concepts, including domain restrictions, critical points, vertical asymptotes, and the quadratic formula. Perfect for students in Grades 10-12.

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