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Use the Vertical Line Test to determine if each graph represents a function. Drag the vertical line across the graph!
Let's explore how to represent the function f(x) = x - 2 in various formats:
The function f(x) = x - 2 with A = {2, 4, 6, 10, 12} can be written as:
{(2,0), (4,2), (6,4), (10,8), (12,10)}
Note: 8 and 10 are not in set B = {0,1,2,4,5,9}!
| x (Input) | f(x) = x - 2 (Output) | In B? |
|---|---|---|
| 2 | 0 | ✅ |
| 4 | 2 | ✅ |
| 6 | 4 | ✅ |
| 10 | 8 | ❌ |
| 12 | 10 | ❌ |
This shows the points (x, f(x)) plotted on a coordinate plane.
| x (Input) | f(x) (Output) |
|---|---|
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 3 |
| 5 | 4 |
One-One (Injective) Proof:
Assume f(a) = f(b), then:
2a - 1 = 2b - 1
⇒ 2a = 2b
⇒ a = b
Thus, f is one-one.
Not Onto (Surjective) Proof:
Consider y = 0 ∈ ℤ (codomain). Is there x ∈ ℤ such that f(x) = 0?
2x - 1 = 0
⇒ x = 0.5 ∉ ℤ
Thus, f is not onto.
One-One (Injective) Proof:
Assume f(a) = f(b), then:
a² + a + 3 = b² + b + 3
⇒ a² - b² + a - b = 0
⇒ (a - b)(a + b + 1) = 0
This gives two cases:
1. a - b = 0 ⇒ a = b (desired result)
2. a + b + 1 = 0 ⇒ b = -a - 1
Now consider a = 0 ⇒ b = -1
f(0) = 0 + 0 + 3 = 3
f(-1) = 1 - 1 + 3 = 3
Correction: The function is actually NOT one-one over ℤ because f(0) = f(-1) = 3.
However, if we restrict to ℕ (natural numbers), it would be one-one.
(i) Find the range of f (ii) Identify the type of function
(i) Range of f:
Calculate f(x) for each x ∈ A:
f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
Thus, range of f = {1, 8, 27, 64}
(ii) Type of function:
• One-One (Injective): Yes, since all outputs are distinct
• Onto (Surjective): No, since not all natural numbers are in the range
Thus, f is injective but not surjective.
(i) f: ℝ → ℝ defined by f(x) = 2x + 1
(ii) f: ℝ → ℝ defined by f(x) = 3 - 4x²
(i) f(x) = 2x + 1
• One-One: Yes, because it's a strictly increasing linear function
• Onto: Yes, because for any y ∈ ℝ, x = (y - 1)/2 gives f(x) = y
Thus, bijective.
(ii) f(x) = 3 - 4x²
• One-One: No, because f(-a) = f(a) for any a (e.g., f(1) = f(-1) = -1)
• Onto: No, because maximum value is 3 (at x=0), so no x gives f(x) > 3
Thus, not bijective.
For f to be onto, both elements of B must be in the range.
We have two cases:
Case 1: f(-1) = 0 and f(1) = 2
-a + b = 0 ⇒ b = a
a + b = 2 ⇒ a + a = 2 ⇒ a = 1, b = 1
Thus, f(x) = x + 1
Case 2: f(-1) = 2 and f(1) = 0
-a + b = 2
a + b = 0 ⇒ b = -a
-a + (-a) = 2 ⇒ -2a = 2 ⇒ a = -1, b = 1
Thus, f(x) = -x + 1
Solutions: (a=1, b=1) or (a=-1, b=1)
f(x) = { 2x + 1 if x > 1 2 if -1 ≤ x ≤ 1 -1 if -3 < x < -1 }
Find: (i) f(3) (ii) f(0) (iii) f(-1.5) (iv) f(2) + f(-2)
(i) f(3): 3 > 1 ⇒ f(3) = 2(3) + 1 = 7
(ii) f(0): -1 ≤ 0 ≤ 1 ⇒ f(0) = 2
(iii) f(-1.5): -3 < -1.5 < -1 ⇒ f(-1.5) = -1
(iv) f(2) + f(-2):
2 > 1 ⇒ f(2) = 2(2) + 1 = 5
-3 < -2 < -1 ⇒ f(-2) = -1
Thus, f(2) + f(-2) = 5 + (-1) = 4
f(x) = { x + 6 if -5 ≤ x < 2 x - 1 if 2 ≤ x < 6 3x - 4 if 6 ≤ x ≤ 9 }
Find: (i) f(-3) + f(2) (ii) f(7) - f(1) (iii) 2f(4) + f(8) (iv) [2f(-2) - f(6)]/[4f(2) + f(-4)]
(i) f(-3) + f(2):
-5 ≤ -3 < 2 ⇒ f(-3) = -3 + 6 = 3
2 ≤ 2 < 6 ⇒ f(2) = 2 - 1 = 1
Total: 3 + 1 = 4
(ii) f(7) - f(1):
6 ≤ 7 ≤ 9 ⇒ f(7) = 3(7) - 4 = 17
-5 ≤ 1 < 2 ⇒ f(1) = 1 + 6 = 7
Total: 17 - 7 = 10
(iii) 2f(4) + f(8):
2 ≤ 4 < 6 ⇒ f(4) = 4 - 1 = 3
6 ≤ 8 ≤ 9 ⇒ f(8) = 3(8) - 4 = 20
Total: 2(3) + 20 = 26
(iv) [2f(-2) - f(6)]/[4f(2) + f(-4)]:
-5 ≤ -2 < 2 ⇒ f(-2) = -2 + 6 = 4
6 ≤ 6 ≤ 9 ⇒ f(6) = 3(6) - 4 = 14
2 ≤ 2 < 6 ⇒ f(2) = 2 - 1 = 1
-5 ≤ -4 < 2 ⇒ f(-4) = -4 + 6 = 2
Numerator: 2(4) - 14 = 8 - 14 = -6
Denominator: 4(1) + 2 = 6
Total: -6/6 = -1
Verify whether S(t) is one-one or not.
S(t) is a quadratic function (parabola).
For t ≥ 0 (realistic time values):
• If a ≥ 0 and g > 0, S(t) is strictly increasing ⇒ one-one
• If a < 0, there might be two times with same distance (before and after peak)
Thus, generally not one-one over all ℝ, but often one-one for t ≥ 0.
The function t(C) = (9/5)C + 32 converts Celsius to Fahrenheit.
(i) t(0) (ii) t(28) (iii) t(-10)
(iv) Find C when t(C) = 212
(v) Find when Celsius = Fahrenheit
(i) t(0): (9/5)(0) + 32 = 32°F
(ii) t(28): (9/5)(28) + 32 = 50.4 + 32 = 82.4°F
(iii) t(-10): (9/5)(-10) + 32 = -18 + 32 = 14°F
(iv) t(C) = 212:
(9/5)C + 32 = 212 ⇒ (9/5)C = 180 ⇒ C = 100°C
(v) C = F:
Set C = (9/5)C + 32 ⇒ C - (9/5)C = 32 ⇒ -(4/5)C = 32 ⇒ C = -40
Thus, -40°C = -40°F