A delivery company assigns each order a unique tracking code. The tracking system uses a function f : ℝ → ℝ defined by f(x) = (3x + 7)/2, where x ∈ ℝ represents the order number and f(x) represents the tracking code.
A delivery company assigns each order a unique tracking code. The tracking system uses a function f : ℝ → ℝ defined by f(x) = (3x + 7)/2.
(i) A logistics analyst claims that no two different orders can ever receive the same tracking code under this system. Verify whether the analyst's claim is correct by checking if f is one-one. [1]
(ii) The system administrator needs every real number to be a valid tracking code output. Check whether f is onto. [1]
(iii) Based on your results in (i) and (ii), determine whether f is invertible. If yes, find f⁻¹(x) and hence find f⁻¹(13). [2]
Show answerHide answer
Let f(a) = f(b) for some a, b ∈ ℝ.
⟹ (3a + 7)/2 = (3b + 7)/2
⟹ 3a + 7 = 3b + 7
⟹ 3a = 3b
⟹ a = b
∴ f is one-one. The analyst's claim is correct.
(ii) Onto (Surjective):
Let y ∈ ℝ (codomain). We need to find x ∈ ℝ such that f(x) = y.
f(x) = y
⟹ (3x + 7)/2 = y
⟹ 3x = 2y − 7
⟹ x = (2y − 7)/3
Since y ∈ ℝ, we have x = (2y − 7)/3 ∈ ℝ, i.e., x ∈ ℝ (domain).
Also, f(x) = f((2y − 7)/3) = (3·(2y − 7)/3 + 7)/2 = (2y − 7 + 7)/2 = 2y/2 = y. ✓
∴ Every y ∈ ℝ has a pre-image in ℝ, so Range = Codomain = ℝ.
∴ f is onto.
(iii) Invertibility and f⁻¹(x):
Since f is both one-one and onto, f is bijective and hence invertible.
Let y = f(x) = (3x + 7)/2.
⟹ 2y = 3x + 7
⟹ x = (2y − 7)/3
∴ f⁻¹(y) = (2y − 7)/3, i.e., f⁻¹(x) = (2x − 7)/3.
Now, f⁻¹(13) = (2(13) − 7)/3 = (26 − 7)/3 = 19/3.
∴ f⁻¹(13) = 19/3.