Worksheet on difference quotients featuring problems to calculate the average rate of change and simplify difference quotients for different functions.

Problem Analysis

The task involves solving problems related to the difference quotient, which is a fundamental concept in calculus used to find the average rate of change of a function over an interval. The difference quotient is given by:

1. For finding the average rate of change from \( x = 1 \) to \( x \):
\[
\frac{f(x) - f(1)}{x - 1}, \quad x \neq 1
\]

2. For finding and simplifying the difference quotient:
\[
\frac{f(x + h) - f(x)}{h}, \quad h \neq 0
\]

We will solve each problem step by step.

---

Problem 1: \( f(x) = 9x \)

#### Step 1: Compute \( f(1) \)
\[
f(1) = 9 \cdot 1 = 9
\]

#### Step 2: Compute \( f(x) - f(1) \)
\[
f(x) - f(1) = 9x - 9
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x) - f(1)}{x - 1} = \frac{9x - 9}{x - 1}
\]

#### Step 4: Simplify the expression
Factor out 9 from the numerator:
\[
\frac{9x - 9}{x - 1} = \frac{9(x - 1)}{x - 1} = 9, \quad \text{for } x \neq 1
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 2: \( f(x) = x^2 - 2x \)

#### Step 1: Compute \( f(1) \)
\[
f(1) = 1^2 - 2 \cdot 1 = 1 - 2 = -1
\]

#### Step 2: Compute \( f(x) - f(1) \)
\[
f(x) - f(1) = (x^2 - 2x) - (-1) = x^2 - 2x + 1
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x) - f(1)}{x - 1} = \frac{x^2 - 2x + 1}{x - 1}
\]

#### Step 4: Simplify the expression
Notice that \( x^2 - 2x + 1 \) is a perfect square:
\[
x^2 - 2x + 1 = (x - 1)^2
\]
Thus:
\[
\frac{x^2 - 2x + 1}{x - 1} = \frac{(x - 1)^2}{x - 1} = x - 1, \quad \text{for } x \neq 1
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 3: \( f(x) = \frac{5}{x + 4} \)

#### Step 1: Compute \( f(1) \)
\[
f(1) = \frac{5}{1 + 4} = \frac{5}{5} = 1
\]

#### Step 2: Compute \( f(x) - f(1) \)
\[
f(x) - f(1) = \frac{5}{x + 4} - 1
\]
Write 1 with a common denominator:
\[
f(x) - f(1) = \frac{5}{x + 4} - \frac{x + 4}{x + 4} = \frac{5 - (x + 4)}{x + 4} = \frac{5 - x - 4}{x + 4} = \frac{1 - x}{x + 4}
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x) - f(1)}{x - 1} = \frac{\frac{1 - x}{x + 4}}{x - 1}
\]
Simplify the complex fraction:
\[
\frac{\frac{1 - x}{x + 4}}{x - 1} = \frac{1 - x}{x + 4} \cdot \frac{1}{x - 1} = \frac{1 - x}{(x + 4)(x - 1)}
\]
Factor out \(-1\) from the numerator \( 1 - x \):
\[
\frac{1 - x}{(x + 4)(x - 1)} = \frac{- (x - 1)}{(x + 4)(x - 1)} = \frac{-1}{x + 4}, \quad \text{for } x \neq 1
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 4: \( f(x) = \sqrt{x + 15} \)

#### Step 1: Compute \( f(1) \)
\[
f(1) = \sqrt{1 + 15} = \sqrt{16} = 4
\]

#### Step 2: Compute \( f(x) - f(1) \)
\[
f(x) - f(1) = \sqrt{x + 15} - 4
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x) - f(1)}{x - 1} = \frac{\sqrt{x + 15} - 4}{x - 1}
\]

#### Step 4: Rationalize the numerator
Multiply the numerator and denominator by the conjugate of the numerator:
\[
\frac{\sqrt{x + 15} - 4}{x - 1} \cdot \frac{\sqrt{x + 15} + 4}{\sqrt{x + 15} + 4} = \frac{(\sqrt{x + 15} - 4)(\sqrt{x + 15} + 4)}{(x - 1)(\sqrt{x + 15} + 4)}
\]
Simplify the numerator using the difference of squares:
\[
(\sqrt{x + 15} - 4)(\sqrt{x + 15} + 4) = (\sqrt{x + 15})^2 - 4^2 = (x + 15) - 16 = x - 1
\]
Thus:
\[
\frac{\sqrt{x + 15} - 4}{x - 1} = \frac{x - 1}{(x - 1)(\sqrt{x + 15} + 4)} = \frac{1}{\sqrt{x + 15} + 4}, \quad \text{for } x \neq 1
\]

#### Final Answer:
\[
\boxed{C}
\]

---

Problem 5: \( f(x) = 9x - 2 \)

#### Step 1: Compute \( f(x + h) \)
\[
f(x + h) = 9(x + h) - 2 = 9x + 9h - 2
\]

#### Step 2: Compute \( f(x + h) - f(x) \)
\[
f(x + h) - f(x) = (9x + 9h - 2) - (9x - 2) = 9x + 9h - 2 - 9x + 2 = 9h
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x + h) - f(x)}{h} = \frac{9h}{h} = 9, \quad \text{for } h \neq 0
\]

#### Final Answer:
\[
\boxed{D}
\]

---

Problem 6: \( f(x) = 4x^2 \)

#### Step 1: Compute \( f(x + h) \)
\[
f(x + h) = 4(x + h)^2 = 4(x^2 + 2xh + h^2) = 4x^2 + 8xh + 4h^2
\]

#### Step 2: Compute \( f(x + h) - f(x) \)
\[
f(x + h) - f(x) = (4x^2 + 8xh + 4h^2) - 4x^2 = 8xh + 4h^2
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x + h) - f(x)}{h} = \frac{8xh + 4h^2}{h} = \frac{8xh}{h} + \frac{4h^2}{h} = 8x + 4h, \quad \text{for } h \neq 0
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 7: \( f(x) = 3 \)

#### Step 1: Compute \( f(x + h) \)
\[
f(x + h) = 3
\]

#### Step 2: Compute \( f(x + h) - f(x) \)
\[
f(x + h) - f(x) = 3 - 3 = 0
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x + h) - f(x)}{h} = \frac{0}{h} = 0, \quad \text{for } h \neq 0
\]

#### Final Answer:
\[
\boxed{B}
\]

---

Problem 8: \( f(x) = \frac{1}{2x} \)

#### Step 1: Compute \( f(x + h) \)
\[
f(x + h) = \frac{1}{2(x + h)}
\]

#### Step 2: Compute \( f(x + h) - f(x) \)
\[
f(x + h) - f(x) = \frac{1}{2(x + h)} - \frac{1}{2x}
\]
Find a common denominator:
\[
f(x + h) - f(x) = \frac{1}{2(x + h)} - \frac{1}{2x} = \frac{x - (x + h)}{2x(x + h)} = \frac{x - x - h}{2x(x + h)} = \frac{-h}{2x(x + h)}
\]

#### Step 3: Compute the difference quotient
\[
\frac{f(x + h) - f(x)}{h} = \frac{\frac{-h}{2x(x + h)}}{h} = \frac{-h}{2x(x + h)} \cdot \frac{1}{h} = \frac{-1}{2x(x + h)}, \quad \text{for } h \neq 0
\]

#### Final Answer:
\[
\boxed{B}
\]

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Final Answers

\[
\boxed{B, B, B, C, D, B, B, B}
\]