Math Problem Statement

Find the functions f ∘ g, g ∘ f, f ∘ f, and g ∘ g and their domains. f(x) = x / (x + 9), g(x) = 7x - 9

Solution

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We are given the functions:

  • f(x)=xx+9f(x) = \frac{x}{x + 9}
  • g(x)=7x9g(x) = 7x - 9

We need to compute the following:

1. Find (fg)(x)(f \circ g)(x):

This means applying g(x)g(x) inside f(x)f(x), i.e., f(g(x))f(g(x)).

(fg)(x)=f(g(x))=f(7x9)(f \circ g)(x) = f(g(x)) = f(7x - 9)

Substitute g(x)=7x9g(x) = 7x - 9 into f(x)f(x):

f(g(x))=7x9(7x9)+9=7x97xf(g(x)) = \frac{7x - 9}{(7x - 9) + 9} = \frac{7x - 9}{7x}

Simplify:

f(g(x))=197xf(g(x)) = 1 - \frac{9}{7x}

2. Domain of (fg)(x)(f \circ g)(x):

The domain of f(g(x))f(g(x)) depends on where the denominator is not zero. So, we solve:

7x0    x07x \neq 0 \implies x \neq 0

Therefore, the domain of fgf \circ g is (,0)(0,)(-\infty, 0) \cup (0, \infty).

3. Find (gf)(x)(g \circ f)(x):

This means applying f(x)f(x) inside g(x)g(x), i.e., g(f(x))g(f(x)).

(gf)(x)=g(f(x))=g(xx+9)(g \circ f)(x) = g(f(x)) = g\left(\frac{x}{x + 9}\right)

Substitute f(x)=xx+9f(x) = \frac{x}{x + 9} into g(x)g(x):

g(xx+9)=7×xx+99=7xx+99g\left(\frac{x}{x + 9}\right) = 7 \times \frac{x}{x + 9} - 9 = \frac{7x}{x + 9} - 9

To simplify:

g(f(x))=7xx+99(x+9)x+9=7x9(x+9)x+9=7x9x81x+9=2x81x+9g(f(x)) = \frac{7x}{x + 9} - \frac{9(x + 9)}{x + 9} = \frac{7x - 9(x + 9)}{x + 9} = \frac{7x - 9x - 81}{x + 9} = \frac{-2x - 81}{x + 9}

4. Domain of (gf)(x)(g \circ f)(x):

The domain of g(f(x))g(f(x)) depends on the domain of f(x)f(x) where f(x)f(x)'s denominator is not zero:

x+90    x9x + 9 \neq 0 \implies x \neq -9

Thus, the domain of gfg \circ f is (,9)(9,)(-\infty, -9) \cup (-9, \infty).

5. Find (ff)(x)(f \circ f)(x):

This means applying f(x)f(x) inside f(x)f(x), i.e., f(f(x))f(f(x)).

(ff)(x)=f(xx+9)(f \circ f)(x) = f\left(\frac{x}{x + 9}\right)

Substitute f(x)f(x) into itself:

f(xx+9)=xx+9xx+9+9f\left(\frac{x}{x + 9}\right) = \frac{\frac{x}{x + 9}}{\frac{x}{x + 9} + 9}

Simplify:

f(f(x))=xx+9x+9(x+9)x+9=xx+9+9(x+9)=x10x+81f(f(x)) = \frac{\frac{x}{x + 9}}{\frac{x + 9( x + 9)}{x + 9}} = \frac{x}{x + 9 + 9(x + 9)} = \frac{x}{10x + 81}

6. Domain of (ff)(x)(f \circ f)(x):

The domain of f(f(x))f(f(x)) is defined where the denominator is not zero:

10x+810    x811010x + 81 \neq 0 \implies x \neq -\frac{81}{10}

Thus, the domain of fff \circ f is (,8110)(8110,)(-\infty, -\frac{81}{10}) \cup (-\frac{81}{10}, \infty).


Would you like further clarification on any of these steps?

Here are 5 related questions:

  1. What happens when we swap the order of composition between two functions?
  2. How would you compute the inverse of f(x)f(x) or g(x)g(x)?
  3. Can we verify the result of a composition by substituting specific values of xx?
  4. What is the significance of domain restrictions in function compositions?
  5. How does the complexity of the function composition affect its graph?

Tip:

Always check the domain restrictions of each individual function before performing composition to avoid undefined operations.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Domain of Functions
Algebra

Formulas

(f ∘ g)(x) = f(g(x))
(g ∘ f)(x) = g(f(x))
Domain restrictions: Denominator ≠ 0

Theorems

Function Composition Theorem
Domain Restriction in Rational Functions

Suitable Grade Level

Grades 10-12