Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
0. Functions
Combining Functions
Problem 1.1.66
Textbook Question
Simplify the difference quotient ƒ(x+h)-ƒ(x)/h
ƒ(x) = 2x² -3x +1

1
Step 1: Start by substituting the function \( f(x) = 2x^2 - 3x + 1 \) into the difference quotient \( \frac{f(x+h) - f(x)}{h} \).
Step 2: Calculate \( f(x+h) \) by replacing \( x \) with \( x+h \) in the function: \( f(x+h) = 2(x+h)^2 - 3(x+h) + 1 \).
Step 3: Expand \( f(x+h) \): \( 2(x+h)^2 = 2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2 \) and \( -3(x+h) = -3x - 3h \). Combine these to get \( f(x+h) = 2x^2 + 4xh + 2h^2 - 3x - 3h + 1 \).
Step 4: Substitute \( f(x+h) \) and \( f(x) \) into the difference quotient: \( \frac{(2x^2 + 4xh + 2h^2 - 3x - 3h + 1) - (2x^2 - 3x + 1)}{h} \).
Step 5: Simplify the expression by canceling out like terms and dividing each term by \( h \). This will give you the simplified form of the difference quotient.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Difference Quotient
The difference quotient is a formula used to find the average rate of change of a function over an interval. It is expressed as (ƒ(x+h) - ƒ(x))/h, where h represents a small change in x. This concept is fundamental in calculus as it leads to the definition of the derivative, which measures the instantaneous rate of change.
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Function Evaluation
Function evaluation involves substituting a specific value into a function to determine its output. In this case, we need to evaluate ƒ(x+h) and ƒ(x) for the given function ƒ(x) = 2x² - 3x + 1. Understanding how to correctly substitute and simplify expressions is crucial for manipulating the difference quotient.
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Algebraic Simplification
Algebraic simplification is the process of reducing expressions to their simplest form. This involves combining like terms, factoring, and canceling common factors. In the context of the difference quotient, simplifying the expression after substituting ƒ(x+h) and ƒ(x) is essential to arrive at a clear and concise result, especially as h approaches zero.
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