Question 1
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If f(x) = 2x + 3 and g(x) = x – 5, what is f○g?

Question 1Select one:

a.
2x - 5


b.
2x – 7


c.
2x – 10


d.
2x - 15


e.
None of these


Question 2
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If f(x) = 2x + 3 and g(x) = x – 5, what is g ○ f?

Question 2Select one:

a.
2x + 1


b.
2x + 2


c.
2x - 1


d.
2x - 2


e.
None of these


Question 3
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Let f(x) =  x2 – 4, x < 0. What is f-1?

Question 3Select one:

a.
-√(x – 4)


b.
-√(x – 4), x > 4


c.
-√(x + 4)


d.
-√(x + 4), x >-4


e.
None of these


Question 4
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If f(x) = 1 / (3x – 5), what is the inverse of f?

Question 4Select one:

a.
1 / (3x + 5)


b.
1 / (3x – 5)


c.
(1/3).(1/x – 5)


d.
(1/3).(1/x + 5) 


e.
None of these


Question 5
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If f(x) = 2x + 3, and g(x) = x2 + x, what is g ○ f?

Question 5Select one:

a.
4x2 + 14x + 12


b.
4x2 + 12x + 9


c.
x2 + 3x + 3


d.
2x2 +2x +3


e.
None of these

Let's break down each question and solve them one by one.

### **Question 1:**
Given \( f(x) = 2x + 3 \) and \( g(x) = x - 5 \), we are asked to find \( (f \circ g)(x) \), which means \( f(g(x)) \).

So, substitute \( g(x) = x - 5 \) into \( f(x) \):
\[
f(g(x)) = f(x - 5) = 2(x - 5) + 3 = 2x - 10 + 3 = 2x - 7
\]
Thus, the answer is:
**b. \( 2x - 7 \)**

---

### **Question 2:**
Here, we need to find \( (g \circ f)(x) \), which means \( g(f(x)) \).

Substitute \( f(x) = 2x + 3 \) into \( g(x) \):
\[
g(f(x)) = g(2x + 3) = (2x + 3) - 5 = 2x + 3 - 5 = 2x - 2
\]
Thus, the answer is:
**d. \( 2x - 2 \)**

---

### **Question 3:**
Let \( f(x) = x^2 - 4 \) for \( x < 0 \). We need to find the inverse \( f^{-1}(x) \).

1. Start by setting \( y = f(x) = x^2 - 4 \).
2. Solve for \( x \):
   \[
   y = x^2 - 4 \implies x^2 = y + 4 \implies x = -\sqrt{y + 4}
   \]
Since \( x < 0 \), we take the negative square root. Therefore:
\[
f^{-1}(x) = -\sqrt{x + 4}
\]
Thus, the answer is:
**d. \( -\sqrt{x + 4}, x > -4 \)**

---

### **Question 4:**
Let \( f(x) = \frac{1}{3x - 5} \). To find the inverse of \( f \), we proceed as follows:

1. Set \( y = \frac{1}{3x - 5} \).
2. Solve for \( x \):
   \[
   y = \frac{1}{3x - 5} \implies 3x - 5 = \frac{1}{y} \implies 3x = \frac{1}{y} + 5 \implies x = \frac{1}{3} \left(\frac{1}{y} + 5\right)
   \]
Therefore, the inverse is:
\[
f^{-1}(x) = \frac{1}{3}\left(\frac{1}{x} + 5\right)
\]
Thus, the answer is:
**d. \( \frac{1}{3}\left(\frac{1}{x} + 5\right) \)**

---

### **Question 5:**
Given \( f(x) = 2x + 3 \) and \( g(x) = x^2 + x \), we are asked to find \( (g \circ f)(x) \), which means \( g(f(x)) \).

Substitute \( f(x) = 2x + 3 \) into \( g(x) \):
\[
g(f(x)) = g(2x + 3) = (2x + 3)^2 + (2x + 3)
\]
First, expand \( (2x + 3)^2 \):
\[
(2x + 3)^2 = 4x^2 + 12x + 9
\]
Now, add \( 2x + 3 \):
\[
g(f(x)) = 4x^2 + 12x + 9 + 2x + 3 = 4x^2 + 14x + 12
\]
Thus, the answer is:
**a. \( 4x^2 + 14x + 12 \)**

---

### Summary of Answers:
1. **b. \( 2x - 7 \)**
2. **d. \( 2x - 2 \)**
3. **d. \( -\sqrt{x + 4}, x > -4 \)**
4. **d. \( \frac{1}{3}\left(\frac{1}{x} + 5\right) \)**
5. **a. \( 4x^2 + 14x + 12 \)**

---

Would you like a more detailed explanation of any of these steps?

Here are 5 related questions:
1. What is the composition of two inverse functions?
2. How do you find the inverse of a quadratic function?
3. How does composition of functions differ from simple function application?
4. How do we determine the domain and range of inverse functions?
5. What is the significance of restricting the domain when finding an inverse?

**Tip:** When dealing with function composition, always substitute carefully and follow the order of operations strictly.

If f(x) = 2x + 3 and g(x) = x – 5, what is f○g?; If f(x) = 2x + 3 and g(x) = x – 5, what is g○f?; Let f(x) = x^2 – 4, x < 0. What is f-1?; If f(x) = 1 / (3x – 5), what is the inverse of f?; If f(x) = 2x + 3, and g(x) = x^2 + x, what is g○f?

Explore how to solve problems involving function composition and inverses with detailed solutions for functions such as f(x) = 2x + 3 and g(x) = x – 5. Learn key algebraic concepts including the composition of functions, inverse functions, and algebraic manipulation, suitable for high school students in Grades 10-12.

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