Math Problem Statement

For each pair of functions, find ( f ◦ g) (x) and (g ◦ f) (x). 11. f(x) = 2x + 1, 12. f(x) = x2 + 3, g(x) = x − 3 g(x) = 2x − 1 13. f(x) = −x2 − 8, g(x) = x2 − 1 14. f(x) = x + 2, g(x) = x − 2

Solution

To solve for (fg)(x)(f \circ g)(x) and (gf)(x)(g \circ f)(x), we need to recall the definition of composition of functions. For two functions f(x)f(x) and g(x)g(x):

  • (fg)(x)=f(g(x))(f \circ g)(x) = f(g(x)), meaning you substitute g(x)g(x) into f(x)f(x).
  • (gf)(x)=g(f(x))(g \circ f)(x) = g(f(x)), meaning you substitute f(x)f(x) into g(x)g(x).

Problem 11:

Given:

  • f(x)=2x+1f(x) = 2x + 1
  • g(x)=x3g(x) = x - 3

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

f(g(x))=f(x3)=2(x3)+1=2x6+1=2x5f(g(x)) = f(x - 3) = 2(x - 3) + 1 = 2x - 6 + 1 = 2x - 5

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

g(f(x))=g(2x+1)=(2x+1)3=2x2g(f(x)) = g(2x + 1) = (2x + 1) - 3 = 2x - 2

Problem 12:

Given:

  • f(x)=x2+3f(x) = x^2 + 3
  • g(x)=2x1g(x) = 2x - 1

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

f(g(x))=f(2x1)=(2x1)2+3=4x24x+1+3=4x24x+4f(g(x)) = f(2x - 1) = (2x - 1)^2 + 3 = 4x^2 - 4x + 1 + 3 = 4x^2 - 4x + 4

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

g(f(x))=g(x2+3)=2(x2+3)1=2x2+61=2x2+5g(f(x)) = g(x^2 + 3) = 2(x^2 + 3) - 1 = 2x^2 + 6 - 1 = 2x^2 + 5

Problem 13:

Given:

  • f(x)=x28f(x) = -x^2 - 8
  • g(x)=x21g(x) = x^2 - 1

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

f(g(x))=f(x21)=(x21)28=(x42x2+1)8=x4+2x218=x4+2x29f(g(x)) = f(x^2 - 1) = -(x^2 - 1)^2 - 8 = -(x^4 - 2x^2 + 1) - 8 = -x^4 + 2x^2 - 1 - 8 = -x^4 + 2x^2 - 9

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

g(f(x))=g(x28)=(x28)21=(x4+16x2+64)1=x4+16x2+63g(f(x)) = g(-x^2 - 8) = (-x^2 - 8)^2 - 1 = (x^4 + 16x^2 + 64) - 1 = x^4 + 16x^2 + 63

Problem 14:

Given:

  • f(x)=x+2f(x) = x + 2
  • g(x)=x2g(x) = x - 2

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

f(g(x))=f(x2)=(x2)+2=xf(g(x)) = f(x - 2) = (x - 2) + 2 = x

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

g(f(x))=g(x+2)=(x+2)2=xg(f(x)) = g(x + 2) = (x + 2) - 2 = x

Summary:

  • Problem 11:
    • (fg)(x)=2x5(f \circ g)(x) = 2x - 5
    • (gf)(x)=2x2(g \circ f)(x) = 2x - 2
  • Problem 12:
    • (fg)(x)=4x24x+4(f \circ g)(x) = 4x^2 - 4x + 4
    • (gf)(x)=2x2+5(g \circ f)(x) = 2x^2 + 5
  • Problem 13:
    • (fg)(x)=x4+2x29(f \circ g)(x) = -x^4 + 2x^2 - 9
    • (gf)(x)=x4+16x2+63(g \circ f)(x) = x^4 + 16x^2 + 63
  • Problem 14:
    • (fg)(x)=x(f \circ g)(x) = x
    • (gf)(x)=x(g \circ f)(x) = x

Would you like further explanation on any of these steps?


Here are five related questions to test your understanding:

  1. What is the composition (fg)(x)(f \circ g)(x) if f(x)=3x+1f(x) = 3x + 1 and g(x)=x24g(x) = x^2 - 4?
  2. How would the result change for (gf)(x)(g \circ f)(x) if f(x)=x2+4f(x) = x^2 + 4 and g(x)=2x+1g(x) = 2x + 1?
  3. Can the composition of two functions ever result in a constant value for all xx?
  4. What is the difference between f(g(x))f(g(x)) and g(f(x))g(f(x)) in terms of domain restrictions?
  5. If f(x)=x3f(x) = x^3 and g(x)=1xg(x) = \frac{1}{x}, what is (fg)(x)(f \circ g)(x)?

Tip: Always ensure that the range of the inner function is within the domain of the outer function when composing functions.

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Math Problem Analysis

Mathematical Concepts

Function Composition
Algebra
Polynomial Functions

Formulas

(f ◦ g)(x) = f(g(x))
(g ◦ f)(x) = g(f(x))

Theorems

Function Composition

Suitable Grade Level

Grades 10-12