Math Problem Statement

Use the pair of functions to find f(g(x)) and g(f(x)). Simplify your answers. Given f(x) = 1/(x - 3) and g(x) = 8/x + 3.

Solution

To solve for f(g(x))f(g(x)) and g(f(x))g(f(x)), we can follow these steps with the provided functions:

  1. Given Functions:
    • f(x)=1x3f(x) = \frac{1}{x - 3}
    • g(x)=8x+3g(x) = \frac{8}{x} + 3

Step 1: Finding f(g(x))f(g(x))

To find f(g(x))f(g(x)), we substitute g(x)g(x) into f(x)f(x): f(g(x))=f(8x+3)f(g(x)) = f\left(\frac{8}{x} + 3\right) Substitute 8x+3\frac{8}{x} + 3 for xx in f(x)=1x3f(x) = \frac{1}{x - 3}: f(g(x))=1(8x+3)3f(g(x)) = \frac{1}{\left(\frac{8}{x} + 3\right) - 3} Simplify inside the denominator: f(g(x))=18x+33=18x=x8f(g(x)) = \frac{1}{\frac{8}{x} + 3 - 3} = \frac{1}{\frac{8}{x}} = \frac{x}{8} So, f(g(x))=x8f(g(x)) = \frac{x}{8}

Step 2: Finding g(f(x))g(f(x))

To find g(f(x))g(f(x)), we substitute f(x)f(x) into g(x)g(x): g(f(x))=g(1x3)g(f(x)) = g\left(\frac{1}{x - 3}\right) Substitute 1x3\frac{1}{x - 3} for xx in g(x)=8x+3g(x) = \frac{8}{x} + 3: g(f(x))=81x3+3g(f(x)) = \frac{8}{\frac{1}{x - 3}} + 3 Simplify the fraction: g(f(x))=8(x3)+3g(f(x)) = 8(x - 3) + 3 Expanding, g(f(x))=8x24+3=8x21g(f(x)) = 8x - 24 + 3 = 8x - 21 So, g(f(x))=8x21g(f(x)) = 8x - 21

Final Answers:

  • f(g(x))=x8f(g(x)) = \frac{x}{8}
  • g(f(x))=8x21g(f(x)) = 8x - 21

Would you like more detailed steps or further explanations?

Related Questions:

  1. How do we find compositions for more complex functions?
  2. What are some common applications of function compositions in real-world problems?
  3. How would we verify if two functions are inverses by using compositions?
  4. Can you explain the concept of domain and range in the context of function compositions?
  5. What happens if we reverse the order in a function composition?

Tip:

When working with compositions, always substitute carefully and simplify step by step to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Algebra

Formulas

f(g(x)) = f(g(x) expression)
g(f(x)) = g(f(x) expression)

Theorems

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Suitable Grade Level

Grades 9-12