Math Problem Statement

For the given functions, find (f ∘ g)(x) and (g ∘ f)(x) and the domain of each. f(x) = (1 - x) / 22x, g(x) = 1 / (1 + 22x)

Solution

We are given the functions f(x)=1x22xf(x) = \frac{1 - x}{22x} and g(x)=11+22xg(x) = \frac{1}{1 + 22x}, and asked to find the compositions (fg)(x)(f \circ g)(x) and (gf)(x)(g \circ f)(x), along with their domains.

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

This means we substitute g(x)g(x) into f(x)f(x):

(fg)(x)=f(g(x))=f(11+22x)(f \circ g)(x) = f(g(x)) = f\left(\frac{1}{1 + 22x}\right)

Now substitute g(x)g(x) into the expression for f(x)f(x):

f(11+22x)=111+22x2211+22xf\left( \frac{1}{1 + 22x} \right) = \frac{1 - \frac{1}{1 + 22x}}{22 \cdot \frac{1}{1 + 22x}}

Simplify the expression:

111+22x=(1+22x)11+22x=22x1+22x1 - \frac{1}{1 + 22x} = \frac{(1 + 22x) - 1}{1 + 22x} = \frac{22x}{1 + 22x} 2211+22x=221+22x22 \cdot \frac{1}{1 + 22x} = \frac{22}{1 + 22x}

Thus,

f(11+22x)=22x1+22x221+22x=xf\left( \frac{1}{1 + 22x} \right) = \frac{\frac{22x}{1 + 22x}}{\frac{22}{1 + 22x}} = x

So, (fg)(x)=x(f \circ g)(x) = x.

2. Find (gf)(x)(g \circ f)(x)

This means we substitute f(x)f(x) into g(x)g(x):

(gf)(x)=g(f(x))=g(1x22x)(g \circ f)(x) = g(f(x)) = g\left( \frac{1 - x}{22x} \right)

Now substitute f(x)f(x) into the expression for g(x)g(x):

g(1x22x)=11+221x22xg\left( \frac{1 - x}{22x} \right) = \frac{1}{1 + 22 \cdot \frac{1 - x}{22x}}

Simplify the denominator:

1+221x22x=1+1xx=x+(1x)x=1x1 + 22 \cdot \frac{1 - x}{22x} = 1 + \frac{1 - x}{x} = \frac{x + (1 - x)}{x} = \frac{1}{x}

Thus,

g(1x22x)=11x=xg\left( \frac{1 - x}{22x} \right) = \frac{1}{\frac{1}{x}} = x

So, (gf)(x)=x(g \circ f)(x) = x.

3. Domains of fgf \circ g and gfg \circ f

  • For (fg)(x)(f \circ g)(x), since both f(x)f(x) and g(x)g(x) simplify to xx, the domain will be all values of xx where g(x)g(x) is defined. The domain of g(x)g(x) is all xx such that 1+22x01 + 22x \neq 0, or x122x \neq -\frac{1}{22}. So, the domain of fgf \circ g is (,122)(122,)(-\infty, -\frac{1}{22}) \cup (-\frac{1}{22}, \infty).

  • For (gf)(x)(g \circ f)(x), similarly, we need f(x)f(x) to be defined, which requires 22x022x \neq 0, or x0x \neq 0. So, the domain of gfg \circ f is (,0)(0,)(-\infty, 0) \cup (0, \infty).

Summary:

  • (fg)(x)=x(f \circ g)(x) = x with domain (,122)(122,)(-\infty, -\frac{1}{22}) \cup (-\frac{1}{22}, \infty)
  • (gf)(x)=x(g \circ f)(x) = x with domain (,0)(0,)(-\infty, 0) \cup (0, \infty)

Would you like more details or have any questions?

Here are some related questions for you:

  1. How do you find the composition of two functions?
  2. What is the domain of a function and how do you determine it?
  3. Why does fgf \circ g simplify to xx in this case?
  4. What are some cases where function compositions do not result in simple identities?
  5. Can the domains of composed functions differ from the domains of the individual functions?

Tip: When finding domains of composite functions, always check the restrictions from both functions in the composition!

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Math Problem Analysis

Mathematical Concepts

Function Composition
Algebra
Domain of Functions

Formulas

f(g(x)) = f(1/(1 + 22x))
g(f(x)) = g((1 - x) / 22x)

Theorems

Function Composition
Domain and Range Theorem

Suitable Grade Level

Grades 10-12