Algebra 1 · Unit 6
Find f(x) for a given x, and find x when you are given f(x), from an equation, graph, or table.
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f(x) is the output
To find f(3), substitute 3 for x. The input is 3, the output is f(3).
Solve f(x) = k
Set the expression equal to k and solve for x. That x is in the domain if the story allows it.
From a graph or table
Read up from x to the graph to get f(x). Read across from a y-value to find x.
Check the story
An x that solves the algebra might still be outside the domain of the story.
Evaluate f(x), or solve f(x) = k, for y = mx + b.
f(3) = 2(3) + 5 = 11
Answer: 11
f(2) = −6 + 1 = −5
Answer: −5
4x − 1 = 11
4x = 12
x = 3
Answer: x = 3
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| f(x) | 5 | 7 | 9 | 11 |
f(3) = 9
Answer: 9
g(0) = −8
x − 8 = 0, x = 8
Answer: g(0) = −8; x = 8
A constant function. Both outputs are 5.
Answer: 5 and 5
A graph, standard form, or a two-step solve.
f(8) = 4 + 3 = 7
12x + 3 = 6
12x = 3
x = 6
Answer: f(8) = 7; x = 6
f(2) = 3
1 = −1x + 5, x = 4
Answer: 3; x = 4
2(3) + 3y = 12
3y = 6
y = 2
f(3) = 2
Answer: 2
C(5) = 45 + 4 = 49
9n + 4 = 49
n = 5
Five tickets cost 49 dollars.
Answer: 49; n = 5
f(3) = 2
f(2) = 6 − 7 = −1
Answer: −1
| x | 0 | 2 | 4 | 6 |
|---|---|---|---|---|
| f(x) | 1 | 5 | 9 | 13 |
When f(x) = 9, x = 4.
Check: each +2 in x adds 4 to f, so m = 2, f(x) = 2x + 1.
Answer: x = 4
A restricted domain, or an input that the story rejects.
V(3) = 40 − 15 = 25
40 − 5t = −5
t = 9
t = 9 is outside 0 ≤ t ≤ 8. The tank is already empty at t = 8. The story has no such t.
Answer: 25; no
f(5) = 93 = 3
2x − 13 = 3
2x − 1 = 9
x = 5
Answer: 3; x = 5
m = 84 = 2
From x = 4 to x = 6 is +2, so y increases by 4: f(6) = 14
f(x) = 2x + 2
f(6) = 12 + 2 = 14
Answer: 14
4 − x ≥ 2
x ≤ 2
x = 0, 1, 2
Answer: 0, 1, 2
2x + 1 = 11
x = 5
A linear function that is not constant is one-to-one, so only x = 5.
Answer: x = 5
1.5(4) + 2 = 6 + 2 = 8. The table is right.
1.5x + 2 = 11
1.5x = 9
x = 6
Answer: x = 6
2m + 3 = 11
m = 4
f(5) = 20 + 3 = 23
Answer: 23
Slope a − bb − a = −1
f(f(a)) = f(b) = a
Answer: slope −1; f(f(a)) = a
F(2) = 11
3h + 5 = 2
3h = −3
h = −1
Negative hours are not in the domain. No.
Answer: 11; no
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