Section 1.1: Functions

Example 1

In the height and age example above, is height a function of age? Is age a function of height?

In the height and age example above, it would be correct to say that height is a function of age, since each age uniquely determines a height. For example, on my 18th birthday, I had exactly one height of 69 inches.

However, age is not a function of height, since one height input might correspond with more than one output age. For example, for an input height of 70 inches, there is more than one output of age since I was 70 inches at the age of 20 and 21.

Example 2

A function \(N = f(y)\) gives the number of police officers, \(N\), in a town in year \(y\). What does \(f(2005) = 300\) tell us?

When we read \(f(2005) = 300\), we see the input quantity is 2005, which is a value for the input quantity of the function, the year (\(y\)). The output value is 300, the number of police officers (\(N\)), a value for the output quantity. Remember \(N=f(y)\). So this tells us that in the year 2005 there were 300 police officers in the town.

Example 3

Which of these tables define a function (if any)?

Input	Output
2	1
5	3
8	6

Input	Output
-3	5
0	1
4	5

Input	Output
1	0
5	2
5	4

The first and second tables define functions. In both, each input corresponds to exactly one output. The third table does not define a function since the input value of 5 corresponds with two different output values.

Example 4

Using the table shown, where \(Q=g(n)\)

\(n\)	1	2	3	4	5
\(Q\)	8	6	7	6	8

a) Evaluate \(g(3)\)
b) Solve \(g(n)=6\)

a) Evaluate \(g(3)\): Evaluating \(g(3)\) (read: g of 3) means that we need to determine the output value, \(Q\), of the function g given the input value of \(n=3\). Looking at the table, we see the output corresponding to \(n=3\) is \(Q=7\), allowing us to conclude \(g(3) = 7\).

b) Solve \(g(n)=6\): Solving \(g(n) = 6\) means we need to determine what input values, \(n\), produce an output value of 6. Looking at the table we see there are two solutions: \(n = 2\) and \(n = 4\). When we input 2 into the function \(g\), our output is \(Q = 6\). When we input 4 into the function \(g\), our output is also \(Q = 6\).

Example 5

Which of these graphs defines a function \(y=f(x)\)?

Looking at the three graphs above, the first two define a function \(y=f(x)\), since for each input value along the horizontal axis there is exactly one output value corresponding, determined by the y-value of the graph. The third graph does not define a function \(y=f(x)\) since some input values, such as \(x=2\), correspond with more than one output value.

Example 6

Given the graph below,

a) Evaluate \(f(2)\).
b) Solve \(f(x) = 4\).

a) To evaluate \(f(2)\), we find the input of \(x=2\) on the horizontal axis. Moving up to the graph gives the point (2, 1), giving an output of \(y=1\). So \(f(2) = 1\).

b) To solve \(f(x) = 4\), we find the value 4 on the vertical axis because if \(f(x) = 4\) then 4 is the output. Moving horizontally across the graph gives two points with the output of 4: (-1,4) and (3,4). These give the two solutions to \(f(x) = 4\): \(x = -1\) or \(x = 3\). This means \(f(-1)=4\) and \(f(3)=4\), or when the input is -1 or 3, the output is 4.

Example 7

Express the relationship \(2n + 6p = 12\) as a function \(p = f(n)\) if possible.

To express the relationship in this form, we need to be able to write the relationship where \(p\) is a function of \(n\), which means writing it as \(p =\) [something involving \(n\)].

\(2n + 6p = 12\)	subtract \(2n\) from both sides
\(6p = 12 - 2n\)	divide both sides by 6 and simplify

\[p=\frac{12-2n}{6}=\frac{12}{6}-\frac{2n}{6}=2-\frac{1}{3}n\]

Having rewritten the formula as \(p=\), we can now express \(p\) as a function: \(p=f(n)=2-\frac{1}{3}n\)

Example 8

Given the function \(k(t)=t^3+2\):
a) Evaluate \(k(2)\).
b) Solve \(k(t)=1\).

a) To evaluate \(k(2)\), we plug in the input value 2 into the formula wherever we see the input variable \(t\), then simplify: \[\begin{align*}k(2)=&2^3+2\\k(2)=&8+2 \end{align*}\] So \(k(2) = 10\).

b) To solve \(k(t) = 1\), we set the formula for \(k(t)\) equal to 1, and solve for the input value that will produce that output: \[\begin{align*} k(t)=&1 & \\ t^3+2=&1 &\text{substitute the original formula} \\ t^3=&-1 &\text{subtract 2 from each side} \\ t=&-1 &\text{take the cube root of each side} \end{align*}\]

When solving an equation using formulas, you can check your answer by using your solution in the original equation to see if your calculated answer is correct.

We want to know if \(k(t) = 1\) is true when \(t=-1\): \[\begin{align*}k(-1)=&(-1)^3+2\\=& -1+2\\=&1,\end{align*}\] which was the desired result.

Example 9

Using the tree table above, determine a reasonable domain and range.

We could combine the data provided with our own experiences and reason to approximate the domain and range of the function \( h = f(c)\). For the domain, possible values for the input circumference c, it doesn’t make sense to have negative values, so \(c > 0\). We could make an educated guess at a maximum reasonable value, or look up that the maximum circumference measured is about 119 feet. With this information we would say a reasonable domain is feet.

Similarly for the range, it doesn’t make sense to have negative heights, and the maximum height of a tree could be looked up to be 379 feet, so a reasonable range is feet.

Example 10

Describe the intervals of values shown on the line graph below using set builder and interval notations:

To describe the values, x, that lie in the intervals shown above we would say, x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5. As an inequality it is \(1\leq x\leq 3\) or \(x \gt 5 \). In interval notation it is \([1,3]\cup(5,\infty)\).

Example 11

Find the domain of each function:
a) \(f(x)=2\sqrt{x+4}\)
b) \(g(x)=\dfrac{3}{6-3x}\)

a) Since we cannot take the square root of a negative number, we need the inside of the square root to be non-negative. \(x+4\geq 0\) when \(x\geq -4\), so the domain of \(f(x)\) is \([-4,\infty)\).

b) We cannot divide by zero, so we need the denominator to be non-zero. \(6-3x=0\) when \(x = 2\), so we must exclude 2 from the domain. The domain of \(g(x)\) is \((-\infty,2)\cup(2,\infty)\).