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


Question 1Select 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


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Question **2**

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 For f(x) = √(x + 5) + 2, what is the domain of f-1?

Question 2Select one:

a.

(-∞, ∞)

b.

(0, ∞)

c.

(-∞, 0)

d.

[2, ∞)

e.

None of these

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Question **3**

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


Question 3Select one:

a.

4x2 + 14x + 12


b.

4x2 + 12x + 9


c.

x2 + 3x + 3


d.

2x2 + 2x + 3


e.

None of these


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Question **4**

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

Question 4Select one:

a.

-√(x – 4)

b.

-√(x – 4), x > 4

c.

-√(x + 4)

d.

-√(x + 4), x >-4

e.

None of these

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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 go through these questions step by step.

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**Question 1**:  
If \( f(x) = \frac{1}{3x - 5} \), what is the inverse of \( f \)?

To find the inverse of a function, we follow these steps:
1. Replace \( f(x) \) with \( y \):  
   \( y = \frac{1}{3x - 5} \)
2. Swap \( x \) and \( y \):  
   \( x = \frac{1}{3y - 5} \)
3. Solve for \( y \):  
   \( x(3y - 5) = 1 \)  
   \( 3y - 5 = \frac{1}{x} \)  
   \( 3y = \frac{1}{x} + 5 \)  
   \( y = \frac{1}{3} \left( \frac{1}{x} + 5 \right) \)

Thus, the inverse function is \( f^{-1}(x) = \frac{1}{3} \left( \frac{1}{x} + 5 \right) \), which corresponds to option **d**.

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**Question 2**:  
For \( f(x) = \sqrt{x + 5} + 2 \), what is the domain of \( f^{-1} \)?

1. The original function has a domain where \( x + 5 \geq 0 \), i.e., \( x \geq -5 \).
2. The range of \( f(x) \) will be \( [2, \infty) \), because the square root term \( \sqrt{x + 5} \) is non-negative and \( +2 \) shifts it up by 2.

Since the range of \( f(x) \) becomes the domain of \( f^{-1}(x) \), the domain of \( f^{-1}(x) \) is **[2, ∞)**, which corresponds to option **d**.

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**Question 3**:  
If \( f(x) = 2x + 3 \), and \( g(x) = x^2 + x \), what is \( f \circ g \)?

1. \( f \circ g \) means \( f(g(x)) \), or applying \( f \) to \( g(x) \).
2. \( g(x) = x^2 + x \), so we substitute this into \( f \):  
   \( f(g(x)) = f(x^2 + x) = 2(x^2 + x) + 3 = 2x^2 + 2x + 3 \)

Thus, \( f \circ g(x) = 2x^2 + 2x + 3 \), which corresponds to option **d**.

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**Question 4**:  
Let \( f(x) = x^2 - 4 \), where \( x < 0 \). What is \( f^{-1} \)?

1. To find the inverse, we start with \( y = x^2 - 4 \).
2. Swap \( x \) and \( y \):  
   \( x = y^2 - 4 \)
3. Solve for \( y \):  
   \( y^2 = x + 4 \)  
   \( y = -\sqrt{x + 4} \), because \( x < 0 \), we choose the negative root.

Thus, the inverse is \( f^{-1}(x) = -\sqrt{x + 4} \), which corresponds to option **c**.

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**Question 5**:  
If \( f(x) = 2x + 3 \), and \( g(x) = x^2 + x \), what is \( g \circ f \)?

1. \( g \circ f \) means \( g(f(x)) \), or applying \( g \) to \( f(x) \).
2. \( f(x) = 2x + 3 \), so we substitute this into \( g \):  
   \( g(f(x)) = g(2x + 3) = (2x + 3)^2 + (2x + 3) \)
3. Expanding the terms:  
   \( (2x + 3)^2 = 4x^2 + 12x + 9 \)  
   \( g(2x + 3) = 4x^2 + 12x + 9 + 2x + 3 = 4x^2 + 14x + 12 \)

Thus, \( g \circ f(x) = 4x^2 + 14x + 12 \), which corresponds to option **a**.

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Do you want further details or have any questions?

### 5 Related Questions:
1. How do you find the inverse of a general linear function?
2. What is the difference between the domain and range of a function?
3. How do composition of functions work in different contexts?
4. Can the inverse of a function always be found, and what determines this?
5. What is the role of the range when determining the domain of an inverse?

**Tip**: Always check if the function is one-to-one before finding the inverse. Functions must be injective to have an inverse.

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

Question 1: Select 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

Learn how to find the inverse of the function f(x) = 1 / (3x – 5) with a detailed step-by-step solution. This problem involves key algebraic concepts related to inverse functions and solving equations. Suitable for students in Grades 9-12.

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