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  1. Math 242: Calculus 1/
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Section 1.1

Four Ways of Representing a Function

Full disclosure: these videos are produced my another professor, and I’m including them here for review.

The following videos are lecture videos covering the material from Section 1.1 in Stewart’s Calculus 8th edition. Each major topic in the section is broken down into a short video.

The following video covers the fundamentals of functions including the basic definition and some visual examples.

There are four basic ways to represent functions and each have their own uses, benefits, and shortcomings.

Next, you are introduced to some basic properties of functions.

Some functions have a property known as symmetry. In the context of functions, you have two main types of symmetry, but later in your mathematical careers you will see that other types of symmetry are possible. In any case, symmetry greatly reduces the complexity of a problem and simplifies our calculations.

Finally, we address some left over properties of functions.

Section 1.1 Exercises

Exercise 1.1.3: Answering questions based on a given graph.

Exercises 1.1.7 & 1.1.8: Determining if a curve is a function

Exercise 1.1.25: Evaluating a function

Exercise 1.1.28: Simplifying the Difference Quotient

Exercise 1.1.32: Finding the Domain of a Function

Working out a question from WebAssign: #

Working out the following question from WebAssign:

WebAssign Question|538x81

The function is $f(x) = \frac1x$, so we begin by writing the difference quotient (the expression at the right - differences and division (quotient)) and then plug everything into the function:

$$ \begin{align*} \dfrac{f(x) - f(a)}{x- a} &= \dfrac{\dfrac1x - \dfrac1a}{x-a} & \\ &= \dfrac{\dfrac{a}{ax} - \dfrac{x}{ax}}{x-a} & \text{getting common denominator up top} \\ &= \dfrac{\dfrac{a-x}{ax}}{x-a} & \\ &= \dfrac{a-x}{ax} \cdot \dfrac{1}{x-a} & \text{reciprocate and multiply} \\ &= \dfrac{-(x-a)}{ax} \cdot\dfrac{1}{x-a} & \text{factor out the negative one} \\ &= \dfrac{-1}{ax} & \text{simplify} \end{align*} $$