Monday, April 23, 2012

Calculus Revisited #1: Functions

Welcome to Calculus Revisited

Greetings. Over the next 21 blog entries (Monday - Friday), Eddie's Math and Calculator Blog will revisit and review some of the major topics in one-variable calculus. I may not get to everything, but I am going to hit a lot of the major topics. Let's start off with some pre-calculus and then head into derivatives and integrals. This blog is going to serve as a reference: whether you are a student or a professional. Enjoy, and let's get started!

Sources Used for this Series

Ross, Debra Anne. "Master Math: Calculus" Career Press: Franklin Lakes, NJ 1998

Silverman, Richard A. Ph.D "Calculus with Analytic Geometry" Prentice-Hall Inc., Englewood Cliffs, NJ 1985


With the preliminairies out of the way...

Function Basics

Function: Is an expression (formula or equation) that associates each element of the domain to a corresponding element of the range. A true function is an expression where every element in the domain corresponds to 1 and only 1 element in a range. So no one element in domain can correspond to two (or more) elements in the range.

Domain: The initial set to which a function is applied. The variable that describes the domain is the independent variable.

Range: The set of results. The variable that describes the range is the dependent variable.

So...

domain → Function → range
xy = f(x)y


Note: For the purpose of this series, I will use x, t, θ (the Greek letter theta) for independent variables and y, f, and g for dependent variables. Obviously any letter or description can be used for variables.

The set of real numbers: that is all numbers including integers, decimals, fractions, and irrational numbers. Real numbers just don't include any complex or imaginary number. So, if √-1 (labeled i or j) is involved, the number is not a real number.

This series will focus exclusively on the set of real numbers.


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1. Find the domain and range of the function y(x) = 2/(x-4).

Note that y(x) is not defined at x=4. Why? y(4) = 2/(4-4) = 2/0. We all know that you can't divide by zero. You can calculate y(x) at various points to verify, however y(x) returns answers that cover the set of real numbers.

Domain: The set of real numbers where x ≠ 4
Range: The set of real numbers

2. Find the domain and range of the function y(x) = √ (16 - x^2).

At first glance, y(x) is defined for all x that make 16 - x^2 nonnegative (positive or zero). To calculate the appropriate domain:

Set:
16 - x^2 ≥ 0
16 ≥ x^2
Take the square root of both sides.
x ≥ -4 and x ≤ 4
The proper domain is
-4 ≤ x ≤ 4

Observe that y(x) hits its maximum point when x = 0
y(0) = √ (16 - 0^2) = √ 16 = 4
Minimum points:
y(4) = √ (16 - 4^2) = 0 and
y(-4) = √ (16 - (-4)^2) = 0

Domain: -4 ≤ x ≤ 4
Range: 0 ≤ y ≤ 4



Composite Functions: Basically, a function of a function. Often noted with a raised circle.

Let f(x) and g(x) be two functions. Then f(g(x)) and g(f(x)) are two composite function. In f(g(x)), the g(x) is treated as the independent variable, and vice versa.

3. Let f(x) = x^2 and g(x) = sin x. Find the composite functions f(g(x)) and g(f(x)).

f(g(x)) = f(sin x) = (sin x)^2 = sin^2 x

Caution: Calculators and most mathematical (if not all) software will not allow the user to enter expressions such as sin^2 x. Use the expression (sin x)^2 instead.

g(f(x)) = g(x^2) = sin (x^2)

Inverse Functions: A function that describes a rule from an element of the range to the element of the domain. If y = f(x), then x = f^-1(y).

4. Let f(x) = 7x + 1. Find the inverse function.

Set y as y = 7x + 1. Solve for x.
y = 7x + 1
y - 1 = 7x
(y - 1)/7 = x

Hence the inverse function is f^-1(y) = (y - 1)/7

Tomorrow: Working with Trigonometric Functions

This blog is property of Edward Shore. © 2012

1 comment:

  1. The linear equations can be solved by graphical method and there are three other methods also.You have described these methods in great manner.
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    ReplyDelete

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