Math Problem Statement

Complete the table below. Assume π‘“ and π‘” are one-to-one functions.

Solution

Let's go through the table step by step to complete it, given that ff and gg are one-to-one functions.

Step 1: Analyze Given Information

  • Given values in the table:
    • f(βˆ’10)=10f(-10) = 10
    • f(βˆ’3)=βˆ’3f(-3) = -3
    • f(10)=βˆ’10f(10) = -10
    • g(βˆ’10)=13g(-10) = 13
    • g(10)=βˆ’10g(10) = -10
    • (f∘g)(10)=βˆ’3(f \circ g)(10) = -3
    • (g∘f)(βˆ’10)=βˆ’10(g \circ f)(-10) = -10
    • (g∘f)(10)=βˆ’3(g \circ f)(10) = -3

Step 2: Use the One-to-One Property

  • Using f(βˆ’10)=10f(-10) = 10:

    • If f(βˆ’10)=10f(-10) = 10, then g(10)=βˆ’10g(10) = -10.
  • Using f(βˆ’3)=βˆ’3f(-3) = -3:

    • If f(βˆ’3)=βˆ’3f(-3) = -3, and gg is one-to-one, then g(βˆ’3)g(-3) should have a unique corresponding value in the set of xx.
  • Using f(10)=βˆ’10f(10) = -10:

    • If f(10)=βˆ’10f(10) = -10, and gg is one-to-one, then g(βˆ’10)=13g(-10) = 13.

Step 3: Complete g(x)g(x) values

  • g(13)=xg(13) = x (This is still unknown, but we can determine this value by seeing what value hasn't been used yet in the set).

Step 4: Fill in Compositions (f∘g)(x)(f \circ g)(x) and (g∘f)(x)(g \circ f)(x)

  • For x=βˆ’10x = -10:

    • (f∘g)(βˆ’10)=f(g(βˆ’10))=f(13)(f \circ g)(-10) = f(g(-10)) = f(13).
    • Since f(13)f(13) is not yet known, leave this blank for now.
    • (g∘f)(βˆ’10)=g(f(βˆ’10))=g(10)=βˆ’10(g \circ f)(-10) = g(f(-10)) = g(10) = -10.
  • For x=βˆ’3x = -3:

    • (f∘g)(βˆ’3)=f(g(βˆ’3))(f \circ g)(-3) = f(g(-3)).
    • (g∘f)(βˆ’3)=g(f(βˆ’3))=g(βˆ’3)(g \circ f)(-3) = g(f(-3)) = g(-3).
  • For x=10x = 10:

    • (f∘g)(10)=f(g(10))=f(βˆ’10)=10(f \circ g)(10) = f(g(10)) = f(-10) = 10.
    • (g∘f)(10)=g(f(10))=g(βˆ’10)=13(g \circ f)(10) = g(f(10)) = g(-10) = 13.

Step 5: Determine Missing Values

Given the one-to-one property:

  • g(13)=βˆ’3g(13) = -3 (because no other values match and f(13)=βˆ’10f(13) = -10).
  • f(13)=βˆ’10f(13) = -10 based on (f∘g)(13)=βˆ’10(f \circ g)(13) = -10.

Final Table:

xxβˆ’10-10βˆ’3-310101313
f(x)f(x)10βˆ’3-3βˆ’10-10-3
g(x)g(x)1310-10-3
f∘g(x)f \circ g(x)13-1010-3
g∘f(x)g \circ f(x)-10-3-313

Would you like to go over any specific part of this table in detail, or do you have any further questions?


Follow-up Questions:

  1. How does the one-to-one property help in solving problems like this?
  2. What are the possible values for f(g(x))f(g(x)) if g(x)=13g(x) = 13?
  3. Can you explain why the compositions (f∘g)(x)(f \circ g)(x) and (g∘f)(x)(g \circ f)(x) might be useful in function analysis?
  4. How would you approach this problem if ff and gg were not one-to-one functions?
  5. What would change in the table if g(13)g(13) had been different?

Tip:

Always cross-check the values by substituting back into the functions, as it can help to verify the correctness of your work.

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Math Problem Analysis

Mathematical Concepts

Functions
Composition of Functions
One-to-One Functions

Formulas

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Theorems

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Suitable Grade Level

High School