Math Problem Statement

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Find the range of the following functions: 1. f(x) = (5x - 1) / (x - 4) 2. f(x) = (2x + 1) / (4x + 8) 3. f(x) = (3x) / (3x - 7)

Solution

You are asking to find the range of the following functions from the image:

  1. f(x)=5x1x4f(x) = \frac{5x - 1}{x - 4}
  2. f(x)=2x+14x+8f(x) = \frac{2x + 1}{4x + 8}
  3. f(x)=3x3x7f(x) = \frac{3x}{3x - 7}

Let's find the range of each function step by step.

1. f(x)=5x1x4f(x) = \frac{5x - 1}{x - 4}

This is a rational function. To find its range:

  • The function will be undefined when the denominator is zero. So, x4=0x - 4 = 0, which gives x=4x = 4.
  • This is a vertical asymptote at x=4x = 4.

Now, check for any horizontal asymptotes:

  • The degrees of the numerator and denominator are the same (both degree 1). The horizontal asymptote is found by taking the ratio of the leading coefficients, which gives: y=51=5y = \frac{5}{1} = 5 Thus, there is a horizontal asymptote at y=5y = 5.

Since the function can approach but never reach y=5y = 5, the range of the function is: (,5)(5,)\boxed{(-\infty, 5) \cup (5, \infty)}

2. f(x)=2x+14x+8f(x) = \frac{2x + 1}{4x + 8}

This can be simplified before finding the range. Factor the denominator: f(x)=2x+14(x+2)f(x) = \frac{2x + 1}{4(x + 2)}

Now, to find the range:

  • The function is undefined when the denominator is zero, so 4(x+2)=04(x + 2) = 0 gives x=2x = -2. There is a vertical asymptote at x=2x = -2.

Next, check the horizontal asymptote:

  • The degrees of the numerator and denominator are both 1 (linear), so the horizontal asymptote is the ratio of the leading coefficients: y=24=12y = \frac{2}{4} = \frac{1}{2} There is a horizontal asymptote at y=12y = \frac{1}{2}.

Thus, the range of the function is: (,12)(12,)\boxed{(-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)}

3. f(x)=3x3x7f(x) = \frac{3x}{3x - 7}

For this rational function:

  • The function is undefined when the denominator is zero, so 3x7=03x - 7 = 0 gives x=73x = \frac{7}{3}. There is a vertical asymptote at x=73x = \frac{7}{3}.

Now, check for horizontal asymptotes:

  • The degrees of the numerator and denominator are the same (both degree 1), so the horizontal asymptote is the ratio of the leading coefficients: y=33=1y = \frac{3}{3} = 1 There is a horizontal asymptote at y=1y = 1.

Therefore, the range of the function is: (,1)(1,)\boxed{(-\infty, 1) \cup (1, \infty)}


Let me know if you need further clarification or details.

Here are five related questions:

  1. How do you find the domain of a rational function?
  2. What is the significance of vertical and horizontal asymptotes in the range of functions?
  3. Can a rational function have oblique asymptotes? If so, how do they affect the range?
  4. What happens to the range if the degree of the numerator is greater than the degree of the denominator?
  5. How does simplifying a rational function help in finding its range?

Tip: Always check for asymptotes when determining the range of rational functions.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Asymptotes
Domain and Range

Formulas

Rational function range: horizontal asymptote and vertical asymptote

Theorems

Horizontal asymptote rule for rational functions with same-degree numerator and denominator
Vertical asymptote (undefined points in denominator)

Suitable Grade Level

Grades 10-12