Math Problem Statement

what is the range f(x)=x^2+2x+1/x^2-x-6

Solution

To find the range of the function f(x)=x2+2x+1x2x6f(x) = \frac{x^2 + 2x + 1}{x^2 - x - 6}, we'll analyze the behavior of the function by looking at its critical points, asymptotes, and the overall structure of the function.

Step 1: Factor the numerator and denominator

First, let's factor both the numerator and the denominator where possible.

  • Numerator: x2+2x+1=(x+1)2x^2 + 2x + 1 = (x + 1)^2
  • Denominator: x2x6x^2 - x - 6

The denominator can be factored as: x2x6=(x3)(x+2)x^2 - x - 6 = (x - 3)(x + 2)

So the function can be rewritten as: f(x)=(x+1)2(x3)(x+2)f(x) = \frac{(x + 1)^2}{(x - 3)(x + 2)}

Step 2: Identify Vertical Asymptotes and Discontinuities

The function has vertical asymptotes or points of discontinuity where the denominator is zero. This occurs at x=3x = 3 and x=2x = -2.

Step 3: Analyze Behavior at Infinity

To determine the behavior of f(x)f(x) as xx approaches infinity or negative infinity, we consider the degrees of the polynomial in the numerator and denominator. Both are degree 2 polynomials.

At large values of xx, the function behaves like: f(x)x2x2=1f(x) \approx \frac{x^2}{x^2} = 1

Step 4: Find Critical Points

To find the critical points, take the derivative of f(x)f(x) and set it equal to zero.

However, analyzing the simplified form: f(x)=1+2x+7(x3)(x+2)f(x) = 1 + \frac{2x + 7}{(x - 3)(x + 2)} we can see that the function equals 1 when 2x+7(x3)(x+2)=0\frac{2x + 7}{(x - 3)(x + 2)} = 0, which never happens because the numerator 2x + 7 is non-zero for all x.

Step 5: Determine Limits at the Asymptotes

Let's check the limits as xx approaches the asymptotes: limx3+f(x)=+,limx3f(x)=\lim_{x \to 3^{+}} f(x) = +\infty, \quad \lim_{x \to 3^{-}} f(x) = -\infty limx2+f(x)=,limx2f(x)=+\lim_{x \to -2^{+}} f(x) = -\infty, \quad \lim_{x \to -2^{-}} f(x) = +\infty

Step 6: Conclusion

Given that the function approaches infinity or negative infinity near the vertical asymptotes and approaches 1 as xx approaches infinity, we conclude that the function's range is:

(,1)(1,+)\boxed{(-\infty, 1) \cup (1, +\infty)}

This range excludes 1, as the function never actually takes the value 1.

Would you like more details or have any questions?

Here are five related questions:

  1. How do you find the domain of a rational function?
  2. What are horizontal and vertical asymptotes, and how are they determined?
  3. How can you use derivatives to find the critical points of a function?
  4. What is the importance of factoring polynomials in rational functions?
  5. How do you analyze the end behavior of a rational function?

Tip: Always check for discontinuities (like vertical asymptotes) when analyzing the range of a rational function.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rational functions
Vertical asymptotes
Limits

Formulas

Quadratic equation (factoring)

Theorems

-

Suitable Grade Level

High School