8.1 Functions

Is height a function of age? Is age a function of height?

Answer: 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.

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

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

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

Answer:  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.

Using the table shown, where [latex]Q=g(n)[/latex]

a) Evaluate [latex]g(3)[/latex].
b) Solve [latex]g(n)=6[/latex].

Answer:

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

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

Which of these graphs defines a function [latex]y=f(x)[/latex]?

Answer:

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

Consider the graph below.

a) Evaluate [latex]f(2)[/latex].

b) Solve [latex]f(x) = 4[/latex].

Answer:

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

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

Express the relationship [latex]2n + 6p = 12[/latex] as a function [latex]p(n)[/latex] if possible.

Answer:

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

[latex]\begin{align*}&2n + 6p = 12\\ &\Rightarrow 6p = 12 - 2n\\ &\Rightarrow p=\frac{12-2n}{6}=\frac{12}{6}-\frac{2n}{6}=2-\frac{1}{3}n \end{align*}[/latex]

Having rewritten the formula to solve for [latex]p[/latex], we can now express [latex]p[/latex] as a function of [latex]n[/latex]:

[latex]p(n)=2-\frac{1}{3}n[/latex]

Given the function [latex]k(t)=t^3+2[/latex]:
a) Evaluate [latex]k(2)[/latex].
b) Solve [latex]k(t)=1[/latex].

Answer:

a) To evaluate [latex]k(2)[/latex], we plug in the input value 2 into the formula wherever we see the input variable [latex]t[/latex], then simplify:
[latex]k(2) = 2^3+2 = 8+2 =10[/latex]

So [latex]k(2) = 10[/latex].

b) To solve [latex]k(t) = 1[/latex], we set the formula for [latex]k(t)[/latex] equal to 1, and solve for the input value that will produce that output:

[latex]\begin{align*} k(t)  &= 1 & \\ &\Rightarrow t^3+2  = 1 &\text{substitute the original formula} \\ &\Rightarrow t^3 = -1 &\text{subtract 2 from each side} \\ &\Rightarrow t = -1 &\text{take the cube root of each side} \end{align*}[/latex]

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 [latex]k(t) = 1[/latex] is true when [latex]t=-1[/latex]:

[latex]\begin{align*}k(-1) & = (-1)^3+2\\ & = -1+2\\ & = 1\end{align*}[/latex]

which was the desired result.

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

Answer:

We could combine the data provided with our own experiences and reason to approximate the domain and range of the function [latex]h = f(c)[/latex]. For the domain, possible values for the input circumference [latex]c[/latex], it doesn’t make sense to have negative values, so [latex]c > 0[/latex]. 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 more than 0 and less than 120 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 more than 0 and less than 380 feet.

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

Answer:

To describe the values of [latex]x[/latex] that lie in the intervals shown above we would say: [latex]x[/latex] is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.

Written as an inequality, it is [latex]1\leq x\leq 3[/latex] or [latex]x \gt 5[/latex].

Written in interval notation, it is [latex][1,3]\cup(5,\infty)[/latex].

Find the domain of each function:

a) [latex]f(x)=2\sqrt{x+4}[/latex]

b) [latex]g(x)=\dfrac{3}{6-3x}[/latex]

Answer:

a) domain? ---> all input values that produce an output ---> all values of [latex]x[/latex] that produce a result using [latex]f(x)=2\sqrt{x+4}[/latex]

Hence we have to analyze how the input is used to produce an output in this case:

To produce [latex]f(x)[/latex] given a value for [latex]x[/latex], we first perform addition on [latex]x[/latex], which never causes an issue.

Then we take a square root, which is an even root, and so we have to make sure that what is underneath the root is greater or equal to zero, i.e., [latex]x+4\ge 0[/latex]. This is because we cannot take an even root of a negative number (note that this is not a problem for an odd root).

Finally, the last operation is multiplication (by 2) and, since multiplication never causes an issue, it does not give us additional restrictions.

Therefore, the only restriction we have on [latex]x[/latex] is that [latex]x+4\geq 0[/latex]. Hence [latex]x\geq -4[/latex], and so the domain of [latex]f(x)[/latex] is [latex][-4,\infty)[/latex].

b) domain? ---> all input values that produce an output ---> all values of [latex]x[/latex] that produce a result using [latex]g(x)=\dfrac{3}{6-3x}[/latex]

Hence we have to analyze how the input is used to produce an output in this case:

To produce [latex]g(x)[/latex] given a value for [latex]x[/latex], we first perform multiplication on [latex]x[/latex] by [latex]3[/latex] ([latex]3x[/latex]), which can be calculated for any value of [latex]x[/latex].

Then we perform subtraction ([latex]6-3x[/latex]), which also never causes problems.

Finally we perform division, which is fine for all values except for division by zero. We cannot get a result if we divide a number by zero and so we have to make sure that the denominator, in this case [latex]6-3x[/latex], is not zero. Hence,

[latex]6-3x\neq 0\Rightarrow -3x=-6\Rightarrow x\neq 2[/latex]

Therefore, the only restriction we have on [latex]x[/latex] is that [latex]x\neq 2[/latex]. Hence, the domain of [latex]g(x)[/latex] is [latex](-\infty,2)\cup(2,\infty)[/latex].

A phone data plan has a basic charge of $30 a month. The plan includes first 2GB free and charges $10 for each additional GB, up to 8 GB total usage, after which it charges $15 for each additional GB. If [latex]d[/latex] is the amount of data used (in GB) and [latex]C[/latex] is the total monthly cost:

a.  Express [latex]C(d)[/latex] as a (piece-wise) formula.
b.  Identify the domain and range of [latex]C(d)[/latex].
c.  Graph [latex]C[/latex] as a function of [latex]d[/latex] for [latex]0\le d\le 10[/latex].
d.  Calculate the cost if 9GB were used.

Answer:

a. Task: [latex]C(d)=?[/latex]

Conditions:

only basic charge for first 2GB [latex]\Rightarrow C(d)=\underbrace{30}_{\text{basic charge}}, 0\le d\le 2[/latex]

after 2GB and up to 8 GB, charge of $10/GB for additional GB
[latex]\Rightarrow C(d)=\underbrace{30}_{\text{cost of first 2GB}}+\underbrace{10(d-2)}_{\text{cost above 2GB}}=10d+10, 2< d\le 8[/latex]

after 8GB, charge of $15/GB for additional GB
[latex]\Rightarrow C(d)=\underbrace{30+10(8-2)}_{\text{cost of first 8GB}}+\underbrace{15(d-8)}_{\text{cost above 8GB}}=15d-30, 2< d\le 8[/latex]

In piece-wise form:

[latex]C(d)=\begin{cases} 30 & 0\le d\le 2 \\ 10d+10& 2< d\le 8 \\ 15d-30 & 8< d \end{cases}[/latex]

b. domain: all possible input values
input: amount of data used [latex]d[/latex]

[latex]C(d)[/latex] can be calculated for any value of [latex]d[/latex] so [latex]d[/latex] can be any real number.

[latex]d=\text{amount of data used}\rightarrow[/latex] can't be negative, no upper limit [latex]\Rightarrow d\ge 0[/latex]

[latex]\Rightarrow domain = [0,\infty)[/latex]

range: all possible output values
output: cost [latex]C(d)[/latex]

cost is at least 30, no upper limit [latex]\Rightarrow C(d)\ge 30[/latex]

[latex]\Rightarrow range = [30,\infty)[/latex]

c. Each section is a linear function so we can graph each section by finding two points that belong to the respective lines:

[latex]\begin{align*} 0\le d\le 2&:C(0)=30; C(2)=30\\ 2< d\le 8&: C(2)=10\cdot 2+10=30; C(8)=10\cdot 8+10=90\\ d>8&:C(8)=15\cdot 8-30=90; C(20)=15\cdot 20-30=270 \end{align*}[/latex]

(Note that for the last section our choice for 20 as input was arbitrary; we could have picked any number greater than 8.)

So the graph looks as follows:

d. [latex]C(9)=?[/latex]

Since [latex]9>8[/latex], [latex]C(d)=15d-30[/latex] applies and so [latex]C(9)=15\cdot 9-30=105[/latex]. Hence the cost if 9GB are used is $105.