IB AA · 3A Concepts of a Function
Evaluate a model, solve it, and notice when it stops being realistic.
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Read the input
t = 0 is the start. Substitute before you reach for a calculator.
Three significant figures
When the power is not a whole number that simplifies, round the decimal to 3 significant figures.
Domain and range still apply
A model can ask for the largest domain or the range of the rule, not only a single output.
A limit of the model
Say what real constraint the formula ignores, such as a top speed, a full pond, or a limited number of customers.
A linear model.
V(t) = 80 − 5t
V(6) = 80 − 5(6)
V(6) = 80 − 30
V(6) = 50
Answer: 50 litres
v(x) = 4x
v(3) = 4(3)
v(3) = 12
Answer: 12 m/s
f(x) = 6x − 4
f(2) = 6(2) − 4
f(2) = 8
Now solve 6x − 4 = 20.
6x = 24
x = 4
Answer: f(2) = 8, x = 4
C = 9 + k4
C = 9 + 204
C = 9 + 5
C = 14
Answer: C = 9 + k4, $14
f(x) = −2x + 8
f(3) = −2(3) + 8
f(3) = 2
Answer: 2
P(m) = 40 + 12m
P(5) = 40 + 12(5)
P(5) = 100
Answer: P(m) = 40 + 12m, 100 pages
A root, a squared denominator, or an exponential.
N = 140 − 45e−0.25t
At t = 0, the power is 0, and e⁰ = 1.
N = 140 − 45
N = 95
At t = 8, N = 140 − 45e−2
N is about 134
A real pond can lose fish to disease or fishing, and it cannot hold a population that the model keeps pushing toward 140 forever.
Answer: 95 at the start, about 134 after 8 months
The denominator is 0 when x − 2 = 0.
The domain is x ≠ 2.
f(4) = 64
f(4) = 32
A square is positive except at x = 2, where f is undefined. The fraction is always positive and can be any positive number.
The range is y > 0.
Answer: x ≠ 2, f(4) = 32, y > 0
f(x) =
The inside of a square root must be at least 0.
3x − 6 ≥ 0
3x ≥ 6
x ≥ 2
The range is y ≥ 0.
= 3
3x − 6 = 9
3x = 15
x = 5
Answer: x ≥ 2, y ≥ 0, x = 5
S = 1.6e0.05 · 4 + 0.7
S = 1.6e0.2 + 0.7
S is about 2.65
The model keeps growing forever. Materials, customers, and fashion will not do that for 40 years.
Answer: about 2.65 million; unlimited growth is not realistic
C = 9 + k4
C = 9 + 204
C = 9 + 5
C = 14
Answer: C = 9 + k4, $14
v(t) = 3.5t
v(4) = 3.5 × 4
v(4) = 14
At 50 seconds the model gives 175 m/s, faster than this drone can fly.
Answer: 14 m/s; the model has no maximum speed
Put a domain, a value, and a limitation together.
f(x) =
The inside of a square root must be at least 0.
3x − 6 ≥ 0
3x ≥ 6
x ≥ 2
The range is y ≥ 0.
= 3
3x − 6 = 9
3x = 15
x = 5
Answer: x ≥ 2, y ≥ 0, x = 5
N = 140 − 45e−0.25t
At t = 0, the power is 0, and e⁰ = 1.
N = 140 − 45
N = 95
At t = 8, N = 140 − 45e−2
N is about 134
A real pond can lose fish to disease or fishing, and it cannot hold a population that the model keeps pushing toward 140 forever.
Answer: 95 at the start, about 134 after 8 months
S = 1.6e0.05 · 4 + 0.7
S = 1.6e0.2 + 0.7
S is about 2.65
The model keeps growing forever. Materials, customers, and fashion will not do that for 40 years.
Answer: about 2.65 million; unlimited growth is not realistic
The denominator is 0 when x − 2 = 0.
The domain is x ≠ 2.
f(4) = 64
f(4) = 32
A square is positive except at x = 2, where f is undefined. The fraction is always positive and can be any positive number.
The range is y > 0.
Answer: x ≠ 2, f(4) = 32, y > 0
f(x) = 54x
f⁻¹(x) = 45x
f⁻¹ turns dollars back into pounds.
Answer: f(x) = 54x, f⁻¹(x) = 45x
v(t) = 3.5t
v(4) = 3.5 × 4
v(4) = 14
At 50 seconds the model gives 175 m/s, faster than this drone can fly.
Answer: 14 m/s; the model has no maximum speed
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