Section 2.4 Algebra with Functions
Example 2.28.
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 2.29.
Suppose \(f(x)= 3x-5\) and \(g(x)\) is given in the graph below:
Evaluate \(7f(2)-(g(-2))^2+1\text{.}\)
\(7f(2)-(g(-2))^2+1=8\)
We start by evaluating the functions, but it doesn't matter which function we start with. Let's just go left-to-right:
From the graph, we can see that \(g(-2)=0\text{.}\) So, we can plug those values into the original expression:
Sometimes you'll have some arithmetic to do inside your function before evaluating it. The key is to work from the inside out.
Example 2.30.
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:
Exercises Practice Problems
Exercise Group.
Suppose \(f(x)= 3x-3\) and \(g(x)\) is given in the graph below:
1.
Evaluate \(2f(2)-g(0)+1\text{.}\)
\(2f(2)-g(0)+1=7\)
2.
Evaluate \(g(2)-(g(-2))^2\text{.}\)
\(g(2)-(g(-2))^2=-12\)
3.
Evaluate \(5g(-1)+4f(3)\text{.}\)
\(5g(-1)+4f(3)=-29\)
4.
Evaluate \(\frac{1}{2}f(5)-2f(0)\text{.}\)
\(\frac{1}{2}f(5)-2f(0)=0\)
5.
Evaluate \((f(-2))^2-f(0)-(g(2))^3\text{.}\)
\((f(-2))^2-f(0)-(g(2))^3=20\)
Exercise Group.
Suppose \(f(x)= x\) and \(g(x)\) is given in the table below:
| x | 1 | 2 | 3 | 4 | 5 |
| g(x) | 6 | 4 | -1 | 7 | 0 |
6.
Evaluate \(2f(2)-g(1)+1\text{.}\)
\(2f(2)-g(1)+1=-1\)
7.
Evaluate \(g(4)-(g(3))^3\text{.}\)
\(g(4)-(g(3))^3=5\)
8.
Evaluate \(2g(1)-3f(5)\text{.}\)
\(2g(1)-3f(5)=-3\)
9.
Evaluate \(\frac{1}{2}f(10)-3f(1)\text{.}\)
\(\frac{1}{2}f(5)-3f(1)=2\)
10.
Evaluate \((f(-2))^2-g(2)-\frac{1}{3}g(5)\text{.}\)
\((f(-2))^2-g(2)-\frac{1}{3}g(5)=0\)
Exercise Group.
Suppose \(g(x)\) is given in the graph below and \(h(x)\) is given in the table below:
| x | 0 | 2 | 4 | 6 | 8 |
| h(x) | 2 | 3 | 9 | 1 | 4 |
11.
Evaluate \(4h(2)-g(0)+3\text{.}\)
\(4h(2)-g(0)+3=5\)
12.
Evaluate \(g(4)-\frac{1}{2}(h(6))^3\text{.}\)
\(g(4)-\frac{1}{2}(h(6))^3=5.5\)
13.
Evaluate \(2g(1)+4h(8)\text{.}\)
\(2g(1)+4h(8)=34\)
14.
Evaluate \(\frac{1}{2}g(10)-3h(0)\text{.}\)
\(\frac{1}{2}g(10)-3h(0)=-2\)
15.
Evaluate \((h(4))^2-6g(2)-\frac{1}{3}g(1)\text{.}\)
\((h(4))^2-6g(2)-\frac{1}{3}g(1)=30\)
Exercise Group.
Suppose \(f(x)\) is given in the table below and \(g(x)=\frac{1}{2}x+2\text{:}\)
| \(x\) | \(f(x)\) |
| \(0\) | \(2\) |
| \(1\) | \(4\) |
| \(2\) | \(6\) |
| \(3\) | \(8\) |
| \(4\) | \(10\) |
16.
Evaluate \(3f(2)-g(2)+3\text{.}\)
\(3f(2)-g(2)+3=18\)
17.
Evaluate \(g(10)-\frac{1}{4}(f(3))^2\text{.}\)
\(g(10)-\frac{1}{4}(f(3))^2=-9\)
18.
Evaluate \(3g(1)+4f(4)\text{.}\)
\(3g(1)+4f(4)=47.5\)
19.
Evaluate \(\frac{1}{5}g(6)-3f(0)\text{.}\)
\(\frac{1}{5}g(6)-3f(0)=-5\)
20.
Evaluate \((f(1))^2-2g(2)-\frac{1}{3}g(-4)\text{.}\)
\((f(1))^2-2g(2)-\frac{1}{3}g(-4)=10\)