1.4: More on Using Functions to Describe Relationships
Focus: Describing the Graphical Behavior of Functions
If a function describing a relationship between two quantities does not have a constant rate of change, then a linear function is not appropriate to model the relationship. In that case, the rate of increase or decrease is getting larger or smaller in size (or magnitude). The function may change from increasing to decreasing. We need to develop more tools to describe this aspect of the function and explore the graphical behavior of the function.
Consider the function \(y=f(x)=x^2 - 4x+5\). Exploring how the values of the input and output change with a table.
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 5 | 2 | 1 | 2 | 5 | 10 |
Notice how the change in the output \(y\) changes for each increase in \(x\) of one.
| Interval | [0,1] | [1,2] | [2,3] | [3,4] | [4,5] |
| \(\Delta x \) | +1 | +1 | +1 | +1 | +1 |
| \(\Delta y \) | -3 | -1 | 1 | 3 | 5 |
| \( \dfrac{\Delta y}{\Delta x} \) | \( \dfrac{-3}{1}=-3 \) | \( \dfrac{-1}{1}=-1 \) | \( \dfrac{1}{1}=1 \) | \( \dfrac{3}{1}=3 \) | \( \dfrac{5}{1}=5 \) |
On the interval \([2, \infty) \), a larger value of x gives a larger value of y. This is how we defined an increasing function in section 1.3. On the graph below, the graph moves upward (for a larger \(y\) ) as we move from left to right (for a larger \(x\) ) when the function is increasing on this interval. On the interval \( (-\infty,2) \), a larger value of x gives a smaller value of y. This is how we defined an decreasing function in section 1.3. On the graph below, the graph moves downward (for a smaller \(y\)) as we move from left to right (for a larger \(x\)) when the function is decreasing on this interval. Be aware that the change in \(x\), \( \Delta x \), does not always have to be \( \Delta x = 1\) as shown on the graph to help clarify the different rates of change.
Key Point: Increasing/Decreasing Functions on a Graph
If a function is increasing on an interval, the graph moves upward from left to right.
If a function is decreasing on an interval, the graph moves upward from left to right.
Moreover, in the introductory example above, notice how the average rates of change on each successive interval get bigger or smaller in size. The rates of increase are getting larger in size on the interval \([2, \infty)\) from \( \dfrac{\Delta y}{\Delta x} = \dfrac{1}{1} \) on [2,3] to \( \dfrac{\Delta y}{\Delta x} = \dfrac{3}{1} \) on [3,4] and \( \dfrac{\Delta y}{\Delta x} = \dfrac{5}{1} \) on [4,5]. Notice how the graph curves upward on the interval \([2, \infty)\) getting steeper and steeper as we move from left to right corresponding to the larger and larger rates of increase as shown on the graph below and the table above. In general, if a function has rates of increase that are getting larger in size over an interval, the shape of that portion of the graph will curve upward from left to right resembling the shape of the graph below.
On the interval \((-\infty, 2)\), the rates of decrease are getting smaller in size, decreasing in size from \( \dfrac{\Delta y}{\Delta x} = \dfrac{-5}{1} \) on [-1,0] to \( \dfrac{\Delta y}{\Delta x} = \dfrac{-3}{1} \) on [0,1] and \( \dfrac{\Delta y}{\Delta x} = \dfrac{-1}{1} \) on [1,2]. The graph moves downward with a smaller and smaller magnitude steepness as we move from left to right along the graph corresponding to the smaller and smaller magnitude rates of decrease as shown in the the graph below and the table above. In general, if a function has rates of decrease that are getting smaller in size over an interval, the shape of that portion of the graph will curve downward from left to right resembling the shape of the graph below.
Consider a second function \(y=g(x)=-(x-2)^2 +5\). Exploring how the values of the input and output change with a table.
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 1 | 4 | 5 | 4 | 1 | -4 |
Notice how the change in the output \(y\) changes for each increase in \(x\) of one.
| Interval | [0,1] | [1,2] | [2,3] | [3,4] | [4,5] |
| \(\Delta x \) | +1 | +1 | +1 | +1 | +1 |
| \(\Delta y \) | +3 | +1 | -1 | -3 | -5 |
| \( \dfrac{\Delta y}{\Delta x} \) | \( \dfrac{3}{1}=3 \) | \( \dfrac{1}{1}=1 \) | \( \dfrac{-1}{1}=-1 \) | \( \dfrac{-3}{1}=-3 \) | \( \dfrac{-5}{1}=-5 \) |
On the interval \((-\infty, 2)\), a larger value of x gives a larger value of y. So the function \(g\) is increasing on \((-\infty, 2)\). On the graph below, notice the graph moves upward (for a larger \(y\) ) as we move from left to right. On the interval \([2, \infty)\), a larger value of x gives a smaller value of y. So the function \(g\) is decreasing on \([2, \infty)\). On the graph below, notice the graph moves downward (for a smaller \(y\) ) as we move from left to right.
On the interval \((-\infty, 2)\), the rates of increase are getting smaller in size from \( \dfrac{\Delta y}{\Delta x} = \dfrac{5}{1} \) on [-1,0] to \( \dfrac{\Delta y}{\Delta x} = \dfrac{3}{1} \) on [0,1] and \( \dfrac{\Delta y}{\Delta x} = \dfrac{1}{1} \) on [1,2]. Notice how the graph curves upward on the interval \((-\infty, 2)\) getting less and less steep as we move from left to right corresponding to the smaller and smaller rates of increase as shown on the graph below and the table above. In general, if a function has rates of increase that are getting smaller over an interval, the shape of that portion of the graph will curve upward from left to right resembling the shape of the graph below.
On the interval \([2, \infty )\), the rates of decrease are getting larger in size from \( \dfrac{\Delta y}{\Delta x} = \dfrac{-1}{1} \) on [2,3] to \( \dfrac{\Delta y}{\Delta x} = \dfrac{-3}{1} \) on [3,4] and \( \dfrac{\Delta y}{\Delta x} = \dfrac{-5}{1} \) on [4,5]. The graph moves downward with a larger and larger magnitude steepness as we move from left to right along the graph corresponding to the larger and larger magnitude rates of decrease as shown in the the graph below and the table above. In general, if a function has rates of decrease that are getting larger in size over an interval, the shape of that portion of the graph will curve downward from left to right resembling the shape of the graph below.
A company's sales over 4 weeks, in thousands of dollars, is shown in the graph below.
a) Are the company's sales increasing or decreasing over the 4 weeks?
b) Describe what it happening to the rate of change in sales over time. Is the rate of increase or decrease getting larger or smaller in size?
Solution
a) The company's sales are increasing over time.
b) The rates of increase in sales in getting smaller as time passes.
The value of the stock of a company reached $600 per share at the start of the year, then began to decrease. The rate of decrease in value became larger and larger in size over the next few months. Draw a graph that illustrates the decrease in value of the stock over this time.
Solution
The graph of the function \(p(t)\) is shown below.
- On what intervals is the function increasing?
- On what intervals is the function decreasing?
- On what interval(s) is the rate of increase in \(f\) getting larger?
- On what interval(s) is the rate of increase in \(f\) getting smaller?
- On what interval(s) is the rate of decrease in \(f\) getting larger in size?
- On what interval(s) is the rate of decrease in \(f\) getting smaller in size?
Solution
- The function is increasing on the intervals \((1, 3)\) and \((4, \infty)\).
- The function is decreasing on the intervals \((-\infty, 1)\) and \((3, 4)\).
- The rate of increase in f is getting larger on \((1,2)\) and \((4, \infty)\).
- The rate of increase in f is getting smaller on \((-2,3)\).
- The rate of decrease in f is getting larger in size on \((3,3.5)\) .
- The rate of decrease in f is getting smaller in size on \((-\infty ,1)\) and \((3.5,4)\).
Note that all values of x are approximated from the graph.
The graph of the function \(y = f(x)\) is given below.
- On what interval(s) is \(f\) f increasing?
- On what interval(s) is \(f\) decreasing?
- On what interval(s) is the rate of increase in \(f\) getting larger?
- On what interval(s) is the rate of increase in \(f\) getting smaller?
- On what interval(s) is the rate of decrease in \(f\) getting larger in size?
- On what interval(s) is the rate of decrease in \(f\) getting smaller in size?
- Answer
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- The function \(f\) increasing \((-2,0)\) and \((3.5, \infty)\).
- The function \(f\) decreasing \((- \infty ,-2)\) and \((0,3.5 )\).
- The rate of increase in f is getting larger on \((-2,-1)\) and \((3.5, \infty)\).
- The rate of increase in f is getting smaller on \((-1,0)\).
- The rate of decrease in f is getting larger in size on \((0,2)\) .
- The rate of decrease in f is getting smaller in size on \((- \infty ,-2)\) and \((2,3.5)\).
Note that all values of x are approximated from the graph.
At the start of the year, a smartphone app had 50,000 monthly users. The number of users increased at smaller and smaller rate until April 1. After that, the number of users began to decrease. From April 1 to June 1, the rate of decrease in users became larger and larger in size over time. From June 1 to September 1, the rate of decrease in users became smaller and smaller in size over time before leveling off at 40,000 users. Sketch a graph for a function \(U = g(t)\) that describes the relationship between the number of users \(U\) in \(t\) months since the start of the year.
- Answer
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Focus: A Problem with Using a Function to Describe a Relationship
Consider the function \(y = f(x) = \dfrac{3}{x-2}\) . For the input \(x = 4\), there is no problem with the corresponding output. The output is a number. \[y = f(4) = \dfrac{3}{4-2}=\dfrac{3}{2} \nonumber\ \]However, for the input \(x = 2\), the output is undefined. \[y = f(2) = \dfrac{3}{2-2}=\dfrac{3}{0} \nonumber\ \] If we are attempting to describe a relationship, such as the value of a car or the height of a ball, an undefined value does not make sense in the context of the application. To address this problem, we must specify which inputs of a function give defined outputs. These specified inputs are the only ones we should use for the function. To help us quantify the inputs and defined outputs of a function, we introduce the concept of the domain and range of a function.
Definition: Domain and Range
The domain of a function is the set of all inputs of a function that give a defined output.
The range of a function is the set of all defined outputs for a given set of inputs.
The function \(y=f(x)\) is described by the table below. Find the domain and range of the function.
| \(x\) | 0 | 1 | 2 | 3 |
| \(y=f(x)\) | 5 | 7 | 6 | 5 |
Solution
The domain is a set or list of all the inputs. The domain of \(f\) is {0,1,2,3}. The range is the set of defined outputs for the given domain. The range of \(f\) is {5,6,7}.
Notice that we did not need to list the output of \(y=5\) twice even though it appeared as the output corresponding to two different inputs. Also, the curly brackets {} tell us that this a list of the included numbers.
The function \(y=g(x)\) is described by the graph below. Find the domain and range of the function.
Solution
On a graph, the inputs are listed on the horizontal axis where there are points with defined values of \(x\) for all values of x in between 1 and 4. Since there are an infinite number of values of \(x\) in this range, we can to use interval notation or an inequality to express this set. The domain of \(g\) is (1,4]. We could also use the inequality \(1 < x \le 4\). Notice that we have included the input \(x=4\) corresponding to the closed circle at this endpoint. However, we have excluded the input \(x=2\) corresponding to the open circle at this endpoint.
On a graph, the outputs are listed on the vertical axis where there are defined values of \(y\) for all values of y in between -1 and 3. The range of \(g\) is [-1,3].
The function \(y=h(x)\) is described by the graph below. Find the domain and range of the function.
Solution
Since the graph opens upward on both ends, the graph continues left and right for larger and smaller values of x than those shown on the graph. Therefore, there are points on the graph for all real values of \(x\) along the horizontal axis. The domain is \( (-\infty , \infty) \).
The smallest output \(y\) along the vertical axis is \(y= -4\). Since the graph opens upward on both ends, the graph continues upward for larger values of \(y\) than those shown on the graph. The range is \( (-4 , \infty) \).
If we are given a formula for a function, such as \(f(x)=x^2+3x\), either a domain will be given or one is implied by all the inputs that give a defined output. Let's explore each numerical operation, the corresponding restrictions entailed to get a defined output, and the effect on the domain with a function that involves each operation.
| Operation | Example | Restrictions with the Operation | Explanation | Domain of the Example Function |
| Addition | \(f(x) = x + 2 \) | none | We can add any number and 2 and get a real number | \(D\) is the set of all real numbers. \[D= (-\infty , \infty) \nonumber \] |
| Subtraction | \(f(x) = x - 2 \) | none | We can subtract any number and 2 and get a real number | \(D\) is the set of all real numbers. \[D= (-\infty , \infty) \nonumber \] |
| Multiplication | \(f(x) = 2x \) | none | We can multiply any number and 2 and get a real number | \(D\) is the set of all real numbers . \[D= (-\infty , \infty) \nonumber \] |
| Division | \(f(x)=\dfrac{2}{x} \) | The denominator \(\neq 0\) | We can't divide 2 by zero and get a real number | \(D\) is the set of all real numbers where \( x \neq 0 \) \[= ( -\infty ,0 ) \cup (0 , \infty) \nonumber \] |
| Power (whole number only) | \(f(x) = x^2 \) | none | We can raise any number to a whole number power | \(D\) is the set of all real numbers \[D = (-\infty , \infty ) \nonumber \] |
| Square & Even Roots | \(f(x) = \sqrt{x} \) | "The input of the radical" \(\ge 0\) (radicand \(\ge 0\) ) | We can't take the square or even root of a negative number | \[D= [0 ,\infty) \nonumber \] |
| Cube & Odd Roots | \(f(x) = \sqrt[3]{x} \) | none | We can take the square or even root of any number | \(D\) is the set of all real numbers \[= (-\infty , \infty) \nonumber \] |
At this point, the only functions with restrictions on the input are those involving division and even roots. For a function with multiple operations, we need to identify the operations in the function and corresponding restrictions on the input. For example, the function \( f(x)= 2x+1 \) involves multiplication and addition neither of which has a restiction. Therefore, the domain is \( (-\infty , \infty ) \).
To determine the domain of a function given by a formula identify the restrictions on the input \(x\) corresponding to each operation in the formula.
Find the domain of the following functions.
- \( f(x)= x^2+3x-2 \)
- \( g(x)= \dfrac{3}{x-2} \)
- \( h(x)= \sqrt{2x-1} \)
Solution
- Since the function \( f(x)= x^2+3x-2 \) involves the operations of addition, subtraction, multiplication, and a power, there are no restrictions on the input. Therefore, the domain of \(f\) is \( (-\infty , \infty ) \)
- The function \( g(x)= \dfrac{3}{x-2} \) involves a subtraction which has no corresponding restrictions. However, the division has the restriction that the denominator \( x - 2 \neq 0\) . Solving for \(x\) , \( x \neq 2\). Therefore, the domain of \(g\) is all real numbers where \( x \neq 2\) or \( (-\infty , 2) \cup (2, \infty )\) in interval notation.
- The only operation in \( h(x)= \sqrt{2x-1} \) that has a restriction is the square root where the input of the radical \( 2x-1 \ge 0\). Solving for x by adding 1 to both sides \[2x \ge 1 \nonumber \] Then dividing by 2 on both sides\[x \ge \dfrac{1}{2} \nonumber \] Therefore the domain is \( [\dfrac{1}{2} , \infty) \).
Now You Try: Exercise \(\PageIndex{3}\)
Find the domain of the following functions.
- \( f(x)= 2x^3+4 \)
- \( g(x)= \dfrac{5x+1}{4x-3} \)
- \( h(x)= \sqrt{2-7x} \)
- \( k(x)= \dfrac{5x+1}{\sqrt{2-7x}} \)
- Answer
-
- The domain of \(f\) is \( (-\infty , \infty ) \)
- The domain of \(g\) is all real numbers where \( x \neq \dfrac{3}{4}\) or \( (-\infty , \dfrac{3}{4}) \cup (\dfrac{3}{4}, \infty )\)
- The domain of \(h\)is \( (-\infty, \dfrac{2}{7}] \).
- The domain of \(k\) is \( (-\infty, \dfrac{2}{7}) \).
Identifying the domain of a function is key to addressing the original problem of undefined outputs by eliminating the inputs whose corresponding outputs are undefined. We could also identify the range if we wanted, but it is usually unnecessary. If we wanted to find the range of the function \( y = f(x)= 2x^3+4 \) exercise 1.4.3a for example, we have two options. We could use the graph to determine the range.
From the graph, the range of outputs \(y\) includes all real numbers on the y-axis. Therefore, the domain of \(f\) is \( (-\infty , \infty ) \). The other option is to use \(f\) to write a formula for the \(x\) if possible, then determine the restrictions on \(y\). Solving for \(y\) by subtracting 4 from both sides \[y -4 = 2x^3 \nonumber \] Then dividing by 2 \[\dfrac{y -4}{2} = x^3 \nonumber \] Then taking the cube root of both sides \[\sqrt[3]{\dfrac{y -4}{2}} = x \nonumber \] This is a formula for \(x\) in terms of \(y\). None of the operations have a restriction. The division is always by 2. Therefore, \(y\) can be any real number and so the range of \(f\) is \( (-\infty , \infty ) \).
Focus: Another Problem with Using Functions to Describe Relationships in Applications
Suppose we used the function \(V=f(t)=40000 - 4000t\) to describe the value \(V\) of a car \(t\) years after it was purchased. Although we could evaluate \(f(-2)\) mathematically, the input \(t=-2\) may not make sense in the context of this application. Two years before the car was purchased may not make sense if the car was brand new when purchased. Although the input \(t=12\) does make sense in the context of this application as \(t=12\) years since it was purchased, the output \(V = f(12)=40000 - 4000(12) = -8000\) does not make sense as the value of a car. The value of a car can't be negative. No one will pay some one else just to have their car (ignoring taxes). A person could argue that the minimum value must be at least a few hundred dollars since it could be sold as scrap metal. Furthermore, it is important to keep in mind the limitations of those models we create. If additional information or data revealed that the linear function that is being used to describe the value of a car is valid for only the first four years, we would need to further restrict the domain accordingly.
If a function is used to model an application, the domain must be restricted so that
- the inputs satisfy all mathematical restrictions
- the inputs make sense as the quantity the input represents in the application
- the inputs have corresponding outputs that make sense as the quantity the output represents in the application
- the inputs reflect any other limitations on the quantities the input and the corresponding output represent.
The function \(V=f(t)=40000 - 4000t\) gives the value \(V\) of a new car \(t\) years after it was purchased years after it was purchased. Find the domain of the function in this application.
Solution
Mathematically, there are no restrictions on the input since the operations in the function are subtraction and multiplication. The input \(t\) makes sense as the years since it was purchased when \(t \ge 0\) since the car was purchased when it was new. The smallest value of the output \(V\) that makes sense as a value of a car is \(V = 0\) since the value can't be negative. Since the value is decreasing over time, the largest value of the car would be \(V=f(0) = 40,000 - 4000(0) = $40,000 \) when the car is brand new. Therefore, the value of the car \(V\) must be in the interval \(0 \le V \le 40,000\). However, the domain is a restriction on the input \(t\) not the output \(V\). We can use the formula for \(V\) to write this double inequality in terms of \(t\), \[0 \le 40000 - 4000t \le 40,000 \nonumber \] Solving for t by subtracting 40,000 from all sides of the double inequality \[-40000 \le -4000t \le 0 \nonumber \] Then dividing by -4000 on all sides of the double inequality and switching the direction of the inequality signs \[10 \ge t \ge 0 \nonumber \] The domain of the application is the interval where \(0 \le t \le 10 \) that satisfies both of these restrictions.
Note that we could have also used the graph of \(V=f(t)\) to approximate the values of t where \(V \ge 0 \) if we were unable to solve the double inequality \(0 \le V \le 40,000\) analytically.
Every function the models a relationship between two quantities in an application should be accompanied by a domain which tells us the range of values for the input for which the model is appropriate and valid.
Example \(\PageIndex{9}\)
The height \(h\) of a ball thrown upward at \( 64 \dfrac{ft}{sec}\) from the ground after t seconds is modeled by the function \(h = s(t) = -16t^2+64t \). Find the domain of the application.
Solution
Mathematically, there are no restrictions on the input since the operations in the function are addition, multiplication, and powers. The input \(t\) makes sense as the seconds since it was thrown when \(t \ge 0\) since a negative value of \(t\), such as \(t = -2\), would before referring to a the time before the ball was thrown. The smallest value of the output \(h\) that makes sense as a height is \(h = 0\) since the ball can't go below the ground. Therefore, we must have \( h \ge 0\). Substituting the formula for \(h\) into this inequality, we have \(h = -16t^2+64t \ge 0\). We will learn later in this course how to solve this type of inequality analytically. In the mean time, we can solve it graphically. On the graph below \( h \ge 0\) when \(0 \le t \le 4 \). The domain of the application is the interval where \(0 \le t \le 4 \) that satisfies both of these restrictions on the input \(t\).
Suppose the cost (in thousands) of removing \(p\) percent of a pollutant from an acre foot of water at a waste water treatment facility is approximated with the model \(C(p)=\dfrac{1.25p+1250}{100-p}\). Find the domain of the application.
Solution
Mathematically, there is a restriction on the input for the division, the denominator \(100 - p \neq 0 \). Solving for p, we have \(p \neq 100 \). The input \(p\) makes sense as the percent removed when \(p \ge 0\) since it is impossible to remove a negative percentage of the pollutant. The largest percent of the pollutant removed is 100% if we remove it all. Therefore, \(0 \le p \le 100 \). The smallest value of the output \(C\) that makes sense is \(C = 0\). Without any additional information, we can't identify an upper bound on the the cost. Therefore, we must have \( C \ge 0\) for the output \(C\) to make sense as a cost. Substituting the formula for \(C\) into this inequality, we have \(\dfrac{1.25p+1250}{100-p} \ge 0\). This is another type of inequality that we will later learn how to solve analytically in this course. In the mean time, we can solve this inequality graphically. On the graph below, \( C \ge 0\) when \( p \le 100 \). The domain of the application is the interval where \(0 \le p < 100 \) that satisfies all three of these restrictions.
Suppose a shoe manufacturer uses test market data to estimate that the number of shoes, N, that they can sell at a particular price p with the function \(N=f(p) = 300 - 0.12p \). Find the domain of the application.
- Answer
-
The domain of the application is the interval where \(0 \le p \le 2,500 \).
Mathematically, there are no restrictions. The input \(p\) makes sense as a price of shoes when \(p \ge 0\) since the price charged can't be negative. The manufacturer can't sell a negative number of shoes, so the output \(N \ge 0\). The values of the input that have a corresponding output that satisfy this restriction are \(p \le = 2,500 \).
A pair of shoes that costs $2,500 is not realistic in most cases. If we had actual test market data on the prices and number of shoes sold, we should restrict the domain even further to a realistic interval of prices for the shoes.
Important Topics of this Section
- Describing varying rates of change on a graph
- Definition of domain
- Definition of range
- Domain and Range from graphs
- Domain and Range from tables
- Domain from a formula
- Domain of an application