Functions and Operations

Subsection A.4 Functions and Operations

Operations, such as addition, subtraction, and multiplication, give us a way of combining numbers and expressions. It is worth the time to read the symbols that tell us what operation is being used.

Functions typically involve operations, but they play a different role. A function is a rule that allows us to assign an output for each allowable input. In algebra, functions are often given by an algebraic expression where we can replace the variable by the input and compute the value of the corresponding output. We also use letters, like \(f\) or \(g\text{,}\) to name our function. The symbol \(f(x)\) stands for the output when function \(f\) is performed on \(x\) and the symbol \(f(5)\) stands for the output when function \(f\) is performed on \(5\text{.}\) When graphing functions, we think of each output of the function as being \(y\) and graph the pairs \((x,f(x))\text{.}\)

Example A.4.27. A Function from Algebra.

Let \(f(x)=5x^2-4x+1\text{.}\) Compute the following:

\(\displaystyle f(0)\)
\(\displaystyle f(2)\)
\(\displaystyle f(t+3)\)

Solution.

a) \(f(0)=0+0+1=1\text{,}\) b) \(f(2)=5(2^2)-4(2)+1=20-8+1=13\text{,}\) c) \(f(t+3)=5(t+3)^2-4(t+3)+1\) which can be simplified to \(f(t+3)=5(t^2+6t+9)-4t-12+1=5t^2+30t+45-4t-11= 5t^2+26t+34\text{.}\)

Functions do not have to have algebraic formulas; for example, we could define a function that assigns the number 0 to all even numbers and 1 to all odd numbers. In this example, \(f(23)=1\text{,}\) \(f(-4)=0\text{,}\) and \(f(18)=0\text{.}\) In geometry, we will be looking at transformations, which are functions that map points to points. Sometimes, we can write geometric functions algebraically like \(f((x,y))=(-x,3y)\) but that goes beyond the scope of this course.

Having seen some examples, we now give the definition of function.

Definition A.4.28.

A function is a mapping that assigns to each input exactly one output.

As noted above the inputs and outputs be numbers or other objects such as points or even lines. What distinguishes a function from other mappings is that each input has exactly one output. Although it is possible for several inputs to have the same output, say \(f(0)=0,f(2)=0,f(4)=0\text{,}\) it is not possible for \(f(0)=0\) and \(f(0)=5\text{.}\) Under the action of a function, each input uniquely determines its output.

Subsubsection A.4.1 Working with squares and square roots

Two functions of interest to us in geometry are the squaring function \(f(x)=x^2\) and the square root function \(g(x)=\sqrt(x)\text{.}\)

Definition A.4.29.

The square of a number \(x\) is the number \(x^2=x\cdot x\text{.}\) Note that the number \(x\) may be negative, positive, or 0, but the square, \(x^2\) is always greater than or equal to 0.

The square root of a number \(x\) is the nonnegative number \(r\) such that \(r^2=x\text{.}\) We use the notation \(\sqrt{x}\) to represent the square root of \(x\text{.}\)

Example A.4.30. Computing Squares and Square Roots.

Compute the following using the definitions of square and square root. A calculator should not be necessary.

\(\displaystyle 5^2\)
\(\displaystyle \sqrt{400}\)
\(\displaystyle (\sqrt{400})^2\)
\(\displaystyle \sqrt{5^2}\)

Answer.

a) \(25\text{,}\) b) \(20\text{,}\) c) \(400\text{,}\) d) \(5\)

In Example A.4.30, we observe how the actions of squaring and taking the square root undo each other. The square of \(5\) is \(25\) and the square root of \(25\) is \(5\text{.}\) Similarly, the square root of \(400\) is \(20\) and the square of \(20\) is \(400\text{.}\) Thus, when we perform a square and a square root in succession, the two actions cancel each other out. For this reason, the squaring function and the square root function are said to be inverse operations on the set of nonnegative numbers

⁶

A number \(x\) is said to be nonnegative if \(x\ge 0\text{.}\)

. Since geometric measurements are nonnegative, we may use the square root function to undo the squaring function and vice versa. Thus, whenever you would like to undo the squaring function in an equation such as in the above example, we will take the square root of both sides.

Principle A.4.31. Inverse Property of Squares and Square Roots.

Let \(x\) be a nonnegative real number. Then \(\sqrt{x^2}=x\) and \((\sqrt{x})^2=x\text{.}\)

Using the Pythagorean Theorem 2.5.2 to find lengths will require squaring and taking square roots of numbers. The equation \(a^2+b^2=c^2\)in the Pythagorean Theorem can be written equivalently as \(\sqrt{a^2 + b^2}=\sqrt{c^2}\) or \(\sqrt{a^2 + b^2}=c\text{.}\) Note the use of the inverse action: \(\sqrt{c^2}=c\text{.}\) In fact, the squaring function also undoes the square root function: \((\sqrt{c})^2=\sqrt{c}\times\sqrt{c}=c\text{.}\) This is what we mean by a pair of inverse functions; applying them one right after the other (in either order) cancels their action. On the other hand, we need to be careful in applying an inverse operation. The square root of \(a^2+b^2\) is not equal to \(a+b\) because the square of \(a+b\) is not \(a^2+b^2\text{.}\) Instead, the square of \(a+b\) is \((a+b)^2=a^2+2ab+b^2\) so that \(\sqrt{a^2+2ab+b^2}=\sqrt{(a+b)^2}=a+b\text{.}\)

Example A.4.32. Finding the Length of a Leg of a Right Triangle.

The length of the hypotenuse is 16 inches and the length of one leg is 10 inches. Determine the length of the other leg.

Solution.

By the Pythagorean Theorem, we know \(10^2+b^2=16^2\) or \(100+b^2=256\text{.}\) Subtracting \(100\) from both sides of the equation, we get \(b^2=156\text{.}\) Taking the square root of both sides of the equation, we see \(b=\sqrt{156}\approx 12.49.\)

Our examples demonstrate that \(\sqrt{x+y}\neq\sqrt{x}+\sqrt{y}\text{,}\) but what if the operation is multiplication or division? In this situation, the expression can be simplified.

Principle A.4.33. Exponents and Roots of Products and Quotients.

Let \(x\) and \(y\) be nonnegative real numbers. Then \((xy)^2=x^2\cdot y^2\) and \(\sqrt{xy}=\sqrt{x}\sqrt{y}.\) If \(y>0\text{,}\) then \((\frac{x}{y})^2=\frac{x^2}{y^2}\) and \(\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\text{.}\)

Moreover, if \(n\) is any nonnegative integer, we can say that \((xy)^n=x^n\cdot y^n\) and \(\sqrt[n]{xy}=\sqrt[n]{x}\sqrt[n]{y}\text{.}\) Also \((\frac{x}{y})^n=\frac{x^n}{y^n}\) and \(\sqrt[n]{\frac{x}{y}}=\frac{\sqrt[n]{x}}{\sqrt[n]{y}}\) when \(y>0\text{.}\)

Subsubsection A.4.2 Other Pairs of Inverse Functions

The square and square root functions are not the only pair of functions that ‘undo’ each other. Some pairs that you may be more familiar with include adding \(4\) and subtracting \(4\) and the two function, dividing by \(10\) and multiplying by \(10\text{.}\) Note how the two functions in each pair undo each other and recall how you have used them to solve equations.

Extending the equality principles, Addition Principle of Equality and Multiplication Principle of Equality, we often use the method of performing the same operation to both sides of an equation. Typically, the chosen operation is the inverse of an operation that appears in the equation.

When working in three dimensional geometry, we will need to compute the cube of a number \(s\) which we can write as \(s^3\) or \(s\times s\times s\text{.}\) The cubing function and cube root function are also inverses of each other.

Example A.4.34. Using Inverses to Solve an Equation.

Solve the equation \(5x^3-4=316\text{.}\)

Solution.

Adding \(4\) to both sides of \(5x^3-4=316\) gives us \(5x^3=320\text{.}\) Next we divide the expressions on both sides by \(5\text{,}\) resulting in \(x^3=64\text{.}\) Finally, we take the cube root of both sides, leaving us with \(x=4\text{.}\) Checking our answer, we verify that \(5(4^3)-4=5(64)-4=320-4=316\) as desired.

Subsubsection A.4.3 Order of Operations

In Example A.4.34, you may have wondered whether it was better to add four first or divide by five. Would it even make a difference?

Because of the way that operations are built on each other

⁷

Multiplication is repeated addition, and exponentiation is repeated multiplication.

, the order of performing algebraic operations is prescribed. When performing computations, we perform exponentiation before multiplication and division and we perform multiplication and division before addition and subtraction. We also perform computations from left to right. Parentheses or other grouping symbols, such as a fraction bar

⁸

We can also use placement and size of type, like \(5^{1+3}\) or \(5x\text{.}\)

, are used when we want to treat a numerical expression as a single quantity. Thus, we start with the innermost parentheses and work our way out.

When there are variables in an expression, we may not be able to perform operations in the standard order. Without a specified value, we cannot add a number like \(5\) to an expression like \(x\text{.}\) Instead, we use the distributive law and concept of like terms to simplify our expression.

As equations become more complex, there is no specific order in which to perform inverse operations. If one or both sides can be simplified, that is often a good first or intermediate step. Undoing operations is typically done in the reverse order, just as we need to unpeel layers of clothing in the evening, each item being removed in the opposite order in which it was put on. Our goal in a linear equation is to get down to a single appearance of the variable and to isolate that variable, as in \(x=\text{.}\) In polynomial equations, we usually aim to manipulate the expression until all variables and nonzero numbers are on one side of the equation so that we can factor the expression and apply Zero Product Property to find all solutions.

Example A.4.35. Solving Equations with Order of Operations.

Solve the equation, \(5(x-3)+8=58\text{.}\)

Solution.

We will demonstrate several methods. To emphasize how order of operations are reversed when undoing operations, we note that adding 8 would be last in a computation. Thus our first step is to subtract 8 (or add \(-8\)) to both sides of our equation, resulting in \(5(x-3)=50\text{.}\) Next we, divide both sides by \(5\text{,}\) leaving \(x-3=10\text{.}\) Finally, we add \(3\) to get \(x=13\text{.}\)

Note that if we start by dividing by 5, Principle A.2.16 tells us that we must also divide 8 by 5. The resulting equation, \(x-3+\frac{8}{5}=\frac{58}{5}\) or \(x-3+1.6=11.6\text{,}\) is correct but more difficult to solve.

Finally, we could solve this equation by first simplifying the left side. Our expression \(5(x-3)+8=5x-15+8=5x-7.\) Replacing the original left side by this simplified version gives \(5x-7=58\text{.}\) Adding \(7\) to both sides gives \(5x=65\) and dividing by 5 gives \(x=13\text{.}\)

Prev Top Next