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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.
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Example 2.24.

Suppose f(x)=x2−7x and we want to evaluate 3f(2)−4. Since we start with functions, we first need to find f(2)=(2)2−7(2)=4−14=−10. We can then plug that into our original expression and simplify: 3(2)−4=3(−10)−4=−34.

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Checkpoint 2.25.

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

Evaluate 7f(2)−(g(−2))2+1.

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:

f(2)=3(2)−5=1

From the graph, we can see that g(−2)=0. So, we can plug those values into the original expression:

7f(2)−(g(−2))2+1=7(1)−(0)2+1=8

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

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Example 2.26.

Suppose f(x)=x2−8 and we want to evaluate −f(2+3)+1. In that case, we start inside the function by simplifying 2+3=5. Then, we can evaluate f(5)=(5)2−8=17. Finally, we can plug that into our original expression to get the final answer:

−f(2+3)+1=−f(5)+1=−(17)+1=−16