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Function Arithmetic: Using Graphs to Evaluate Various Combinations of Functions at a Given Value

Function Arithmetic Using Graphs

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.)

Note that \((f+g)(1)=f(1)+g(1)\) so we need to find \(f(1)\) and \(g(1)\) first. \(f\) is the green wiggly function, and the \(y\)-value of \(f\) at \(x=1\) is about \(1.25\). Similarly, the \(y\)-value of the blue function \(g\) at \(x=1\) is \(2.75\)ish. So, we can estimate the value of \((f+g)(1)\) as

$$\begin{align}(f+g)(1)&=f(1)+g(1)\\&=1.25+2.75\\&=4\end{align}$$

Similarly, to compute \((f/g)(2)=\frac{f(2)}{g(2)}\) we first need to compute \(f(2)\) and \(g(2)\) using the graph. Note that at \(x=2\) the green function \(f\) has \(y\)-value roughly \(0.5)\) and the blue function \(g\) has \(y\)-value roughly equal to \(1.25\). So, \((f/g)(2)\) is therefore

$$\begin{align}\left(\frac{f}{g}\right)(2)&=\frac{f(2)}{g(2)}\\&=\frac{0.5}{1.25}\\&=0.4\end{align}$$

Overall, performing function arithmetic using graphs (or any other means, really) involves nothing more than computing each function’s \(y\)-value at the given \(x\) value, and then combining the \(y\)-values as indicated by the operation, such as \(+\), \(x\), \(/\), etc.

Note that the following answers are found by estimating \(y\)-values from a drawn graph.

\((f+g)(2)=f(2)+g(2)=2.5+2=4.5\)

\((g-f)(0)=g(0)-f(0)=0-1=-1\)

\((fg)(-1)=f(-1)\cdot g(-1)=(-1)\cdot (2)=-2\)

In all cases, find the \(y\)-values at the given \(x\)-value, then either add, subtract, multiply, etc. based on what the problem is asking you to do.

Note that the following answers are based on estimating \(y\)-values of the functions shown.

\((f+g)(1)=f(1)+g(1)=(1.25)+(-2)=-0.75\)

\((f-g)(0)=f(0)-g(0)=(-2)-(-1)=-1\)

\((fg)(3)=(f(3))(g(3))=(1/2)\cdot (-4)=-2\)

\(\left(\frac{g}{f}\right)(-3)=\left(\frac{g(-3)}{f(-3)}\right)=\frac{(-4)}{2}=-2\)

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