Function Arithmetic: Using Tables to Evaluate Various Combinations of Functions at a Given Value
Function Arithmetic Using Tables
Recall that when you are given an expression such as \((f+g)(2)\), this means “find the outputs of \(f\) and \(g\) on input \(2\), then add the outputs.” More succinctly, \((f+g)(2)=f(2)+g(2)\). This interpretation, along with how to evaluate the expression, works exactly the same way regardless of how your functions are represented (i.e. its the same if your functions are defined by a graph, table, set, rule, machine, etc.)
| \(x\) | \(f(x)\) | \(g(x)\) |
| 3.2 | 4 | 7 |
| 1 | 3 | 7.2 |
| 0 | -4.23 | 12.15 |
| 2 | 9.4 | -3 |
| 5.214 | -11 | 4 |
To compute \((fg)(1)\), notice first that \((fg)(1)=f(1)\cdot g(1)\). Thus, we need to compute \(f(1)\) and \(g(1)\) and multiply the results.
From the table, reading across the row that starts with \(x=1\), we get \(f(1)=3\) and \(g(1)=7.2\). So, \(f(1)\cdot g(1)=3\cdot 7.2=21.6\)
To compute \((f-g)(2)\), notice that \((f-g)(2)=f(2)-g(2)\). So, we need to compute \(f(2)\) and \(g(2)\) and subtract the results.
From the table, reading across the row that starts with \(x=2\), we get \(f(2)=9.4\) and \(g(2)=-3\). So, we have \((f-g)(2)=f(2)-g(2)=9.4-(-3)=12.4\). Done.
Generalizing a Bit…
In general, when given a combination of functions such as \((fg)(x)\) or \((f-g)(x)\) (for example), each of these is telling you to evaluate each function on the same input value \(x\), and then perform the given operation on the outputs (e.g. multiplication or subtraction, in this case).
When dealing with a variable \(x\) instead of a concrete number (such as \(x=1\) or \(x=2\) in the previous examples above), this \(x\) represents any possible input value you could plug in to your functions. When dealing with tables, all possible input values are listed in the leftmost column. The outputs of the given functions on each of those \(x\)-value inputs are shown in the rightmost columns.
With that information, it is possible to determine what the values of \((f+g)(x)\), \((f-g)(x)\), etc. are based on each of the inputs in the leftmost column of your table. This is exhibited in the example below.
| \(x\) | \(f(x)\) | \(g(x)\) |
| 3.2 | 4 | 7 |
| 1 | 3 | 7 |
| 0 | -4.23 | 12 |
| 2 | 9 | -3 |
| 30 | -11 | 4 |
Essentially, what this problem is asking us to do is compute the value of \((f-g)(x)\) and \((fg)(x)\) for each \(x\) value where both \(f\) and \(g\) are defined. This means building new columns for our table: one for \((f-g)(x)\) and one for \((fg)(x)\). Viz:
| \(x\) | \(f(x)\) | \(g(x)\) | \((f-g)(x)\) | \((fg)(x)\) |
| -1 | 4 | 7 | ||
| 1 | 3 | 7 | ||
| 0 | -4 | 12 | ||
| 2 | 9 | -3 | ||
| 30 | -11 | 4 |
Your goal now is to fill in the last two columns based on what you see in the left three columns.
Recall that \((f-g)(x)=f(x)-g(x)\) and \((fg)(x)=f(x)\cdot g(x)\). So, to fill out the last two columns, all you need to do is compute \(f(x)\) and \(g(x)\) for each \(x\)-value in the left column, and either subtract or multiply the results. This is shown below.
| \(x\) | \(f(x)\) | \(g(x)\) | \((f-g)(x)\) | \((fg)(x)\) |
| -1 | 4 | 7 | \(4-7=-3\) | \(4\cdot 7\) |
| 1 | 3 | 7 | \(3-7=-4\) | \(3\cdot 7\) |
| 0 | -4 | 12 | \(-4-12=-16\) | \(-4\cdot 12\) |
| 2 | 9 | -3 | \(9-(-3)=12\) | \(9\cdot (-3)\) |
| 30 | -11 | 4 | \(-11-4=-15\) | \((-11)\cdot 4=-44\) |
| x | f(x) | g(x) |
| -3 | 5 | 2 |
| 1 | 6 | 7 |
| 2 | 3 | -81 |
| 3 | -7 | -3.14 |
| 4 | 3 | 1 |
| 56 | 1 | 1 |
\((f+g)(2)=f(2)+g(2)=3+(-81)=-78
\((fg)(1)=f(1)\cdot g(1)=6\cdot 7=42
\((f/g)(1)=f(1)/g(1)=3/1=3
Recall that \((f+g)(2)\) means “find the output of \(f\) and \(g\) on input 2, then add the results/outputs.” Use the table to find the outputs/\(y\)-values associated with the input of \(2\) and then add them. Similar reasoning can be used for the other two parts of the problem.
| x | f(x) | g(x) |
| -3 | 7 | 3 |
| 0 | -2 | -5 |
| 1 | -7 | 6 |
| 3 | 3 | 12 |
| 2 | 15 | 7 |
\((f+g)(3)=f(3)+g(3)=3+12=15\)
\((fg)(0)=f(0)\cdot g(0)=(-2)(-5)=10\)
\((f-g)(2)=f(2)-g(2)=15-7=8\)
Recall that \((f+g)(3)\) means “find the output of \(f\) and \(g\) on input 3, then add the results/outputs.” Use the table to find the outputs/\(y\)-values associated with the input of \(2\) and then add them. Similar reasoning can be used for the other two parts of the problem.
| x | f(x) | g(x) |
| -3 | 7 | 3 |
| 0 | -2 | -5 |
| 1 | -7 | 6 |
| 3 | 3 | 12 |
| 2 | 15 | 7 |
| \(x\) | \(f(x)\) | \(g(x)\) | \((f+g)(x)\) | \((fg)(x)\) |
| -3 | 7 | 3 | 10 | 21 |
| 0 | -2 | -5 | -7 | 10 |
| 1 | -7 | 6 | -1 | -42 |
| 3 | 3 | 12 | 15 | 36 |
| 2 | 15 | 7 | 22 | 105 |
To compute each value of the column \((f+g)(x)\), note that \((f+g)(x)=f(x)+g(x)\). So, plug in each \(x\)-value and add the results from both \(f\) and \(g\). Similarly for \((fg)(x)=f(x)\cdot g(x)\); multiply the results instead. Viz:
| \(x\) | \(f(x)\) | \(g(x)\) | \((f+g)(x)\) | \((fg)(x)\) |
| -3 | 7 | 3 | \(f(x)+g(x)=7+3=10\) | \(f(x)\cdot g(x)=7\cdot 3=21\) |
| 0 | -2 | -5 | \(f(x)+g(x)= -2+-5=-7\) | \(f(x)\cdot g(x)= -2\cdot -5=10\) |
| 1 | -7 | 6 | \(f(x)+g(x)= -7+6=-1\) | \(f(x)\cdot g(x)= -7\cdot 6=-42\) |
| 3 | 3 | 12 | \(f(x)+g(x)= 3+12=15\) | \(f(x)\cdot g(x)= 3\cdot 12=36\) |
| 2 | 15 | 7 | \(f(x)+g(x)= 15+7=22\) | \(f(x)\cdot g(x)= 15\cdot 7=105\) |