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The images you've uploaded contain a list of problems and the corresponding functions required to solve them. I'll first outline the functions provided in the second image and then solve the problems listed in the first image.

### Functions Provided:
1. \( f(x) = 2x + 1 \)
2. \( g(x) = 3x - 5 \)
3. \( h(x) = \sqrt{x + 1} \)

### Problems to Solve:
Given these functions, you need to find the following values:

1. \( (g \circ f)(x) \)
2. \( (g \circ g)(x) \)
3. \( (f \circ g)(x) \)
4. \( (g \circ h)(x) \)
5. \( (g \circ h)(2) \)
6. \( (g \circ f)(2) \)
7. \( (f \circ g)(-3) \)
8. \( (h \circ g)\left(\frac{1}{4}\right) \)
9. \( (h \circ f)(-1) \)
10. \( (h \circ g)\left(\frac{4}{3}\right) \)

### Solutions:

1. **\( (g \circ f)(x) \)**
   \[
   g(f(x)) = g(2x + 1) = 3(2x + 1) - 5 = 6x + 3 - 5 = 6x - 2
   \]

2. **\( (g \circ g)(x) \)**
   \[
   g(g(x)) = g(3x - 5) = 3(3x - 5) - 5 = 9x - 15 - 5 = 9x - 20
   \]

3. **\( (f \circ g)(x) \)**
   \[
   f(g(x)) = f(3x - 5) = 2(3x - 5) + 1 = 6x - 10 + 1 = 6x - 9
   \]

4. **\( (g \circ h)(x) \)**
   \[
   g(h(x)) = g(\sqrt{x + 1}) = 3\sqrt{x + 1} - 5
   \]

5. **\( (g \circ h)(2) \)**
   \[
   g(h(2)) = g(\sqrt{2 + 1}) = g(\sqrt{3}) = 3\sqrt{3} - 5
   \]

6. **\( (g \circ f)(2) \)**
   \[
   g(f(2)) = g(2(2) + 1) = g(5) = 3(5) - 5 = 15 - 5 = 10
   \]

7. **\( (f \circ g)(-3) \)**
   \[
   f(g(-3)) = f(3(-3) - 5) = f(-9 - 5) = f(-14) = 2(-14) + 1 = -28 + 1 = -27
   \]

8. **\( (h \circ g)\left(\frac{1}{4}\right) \)**
   \[
   h(g\left(\frac{1}{4}\right)) = h\left(3\left(\frac{1}{4}\right) - 5\right) = h\left(\frac{3}{4} - 5\right) = h\left(\frac{-17}{4}\right)
   \]
   Since the expression inside \( h(x) \) results in a negative value, the square root will involve imaginary numbers. Thus:
   \[
   = \sqrt{\left(\frac{-13}{4}\right) + 1} = \sqrt{\frac{-13 + 4}{4}} = \sqrt{\frac{-9}{4}} = \frac{i\sqrt{9}}{2} = \frac{3i}{2}
   \]

9. **\( (h \circ f)(-1) \)**
   \[
   h(f(-1)) = h(2(-1) + 1) = h(-2 + 1) = h(-1) = \sqrt{-1 + 1} = \sqrt{0} = 0
   \]

10. **\( (h \circ g)\left(\frac{4}{3}\right) \)**
    \[
    h\left(g\left(\frac{4}{3}\right)\right) = h\left(3\left(\frac{4}{3}\right) - 5\right) = h(4 - 5) = h(-1) = \sqrt{-1 + 1} = \sqrt{0} = 0
    \]

Would you like any further details or explanations on these solutions? 

Here are some related questions to consider:
1. What happens if \( g(x) \) or \( f(x) \) involve higher-order polynomials?
2. How do compositions of functions affect the domain and range?
3. Can the composition of two functions be commutative, i.e., \( f(g(x)) = g(f(x)) \)?
4. What are the potential issues when dealing with square roots in functions?
5. How do you interpret the meaning of the composition of functions in a real-world context?

**Tip:** Always check the domain of the inner function first when performing function compositions, as it will affect the domain of the entire expression.

Explore detailed solutions to composition of functions involving square roots. Learn how to solve problems using functions like g(x) = 3x - 5 and h(x) = sqrt(x + 1). Suitable for Grades 11-12 mathematics students.

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