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The image contains two mathematical problems, each dealing with composition of functions.

### Problem 1:
Functions:
- \( f(x) = 2x - 1 \)
- \( g(x) = x + 3 \)

You're tasked with finding \( g(f(x)) \), which is the composition of the functions \( g \circ f \).

Steps:
1. \( g(f(x)) = g(2x - 1) \)
2. Substitute \( 2x - 1 \) into \( g(x) \): 
   \[
   g(2x - 1) = (2x - 1) + 3 = 2x + 2
   \]

Then there are specific values for \( g(f(-1)) \) and \( g(f(-5)) \):
- \( g(f(-1)) \):  
  1. \( f(-1) = 2(-1) - 1 = -3 \)
  2. \( g(-3) = -3 + 3 = 0 \)
  
- \( g(f(-5)) \):  
  1. \( f(-5) = 2(-5) - 1 = -11 \)
  2. \( g(-11) = -11 + 3 = -8 \)

### Problem 2:
Functions:
- \( f(x) = x^2 + 14 \)
- \( g(x) = 5x + 3 \)

You are again tasked with finding the composition of functions \( g(f(x)) \).

Steps:
1. \( g(f(x)) = g(x^2 + 14) \)
2. Substitute \( x^2 + 14 \) into \( g(x) \):
   \[
   g(x^2 + 14) = 5(x^2 + 14) + 3 = 5x^2 + 70 + 3 = 5x^2 + 73
   \]

For specific values:
- \( g(f(8)) \):
  1. \( f(8) = 8^2 + 14 = 64 + 14 = 78 \)
  2. \( g(78) = 5(78) + 3 = 390 + 3 = 393 \)

- \( g(f(6)) \):
  1. \( f(6) = 6^2 + 14 = 36 + 14 = 50 \)
  2. \( g(50) = 5(50) + 3 = 250 + 3 = 253 \)

Would you like further clarification or more details on any of these steps?

### Relative Questions:
1. How would you find \( f(g(x)) \) instead of \( g(f(x)) \)?
2. What happens if you change the functions slightly, e.g., \( f(x) = x^2 - 5x \) and \( g(x) = 2x + 1 \)?
3. Can you find the derivative of the composition \( g(f(x)) \) if \( f(x) = 3x^2 \) and \( g(x) = \sin(x) \)?
4. What are the domain and range of \( g(f(x)) \) for these specific functions?
5. How would you solve for \( x \) if you are given \( g(f(x)) = 0 \)?

### Tip:
When dealing with compositions, always start by substituting the inner function into the outer function carefully, and work step by step to avoid mistakes.

Find the composition of functions f(x) = 2x - 1 and g(x) = x + 3; and f(x) = x^2 + 14 and g(x) = 5x + 3. Calculate g(f(x)) for specific values.

Learn how to solve function composition problems with step-by-step solutions for f(x) = 2x - 1 and g(x) = x + 3. Suitable for students in Grades 10-12. This example includes specific value calculations like g(f(-1)) and g(f(-5)), and covers algebraic concepts.

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