Algebra with Functions

Section 2.4 Algebra with Functions

Functions are just a kind of group, like a fancy version of parentheses. So, if functions show up in more complicated expressions, you evaluate them first.

Example 2.24.

Suppose \(f(x)=x^2-7x\) and we want to evaluate \(3f(2)-4\text{.}\) Since we start with functions, we first need to find \(f(2)=(2)^2-7(2) = 4-14=-10\text{.}\) We can then plug that into our original expression and simplify: \(3(2)-4 = 3(-10)-4 = -34\text{.}\)

Checkpoint 2.25.

Suppose \(f(x)= 3x-5\) and \(g(x)\) is given in the graph below:

Evaluate \(7f(2)-(g(-2))^2+1\text{.}\)

Answer.

\(7f(2)-(g(-2))^2+1=8\)

Solution.

We start by evaluating the functions, but it doesn't matter which function we start with. Let's just go left-to-right:

\begin{equation*} f(2)=3(2)-5 = 1 \end{equation*}

From the graph, we can see that \(g(-2)=0\text{.}\) So, we can plug those values into the original expression:

\begin{equation*} 7f(2)-(g(-2))^2+1 = 7(1)-(0)^2+1=8 \end{equation*}

Sometimes you'll have some arithmetic to do inside your function before evaluating it. The key is to work from the inside out.

Example 2.26.

Suppose \(f(x)=x^2-8\) and we want to evaluate \(-f(2+3)+1\text{.}\) In that case, we start inside the function by simplifying \(2+3=5\text{.}\) Then, we can evaluate \(f(5)=(5)^2-8=17\text{.}\) Finally, we can plug that into our original expression to get the final answer:

\begin{equation*} -f(2+3)+1 = -f(5)+1=-(17)+1=-16 \end{equation*}