Which statement describes function composition with respect to the commutative property? Given f(x) = x² – 4 and g(x) = x – 3, (f ∘ g)(2) = –3 and (g ∘ f)(2) = –3, so function composition is commutative. Given f(x) = 2x – 5 and g(x) = 0.5x – 2.5, (f ∘ g)(x) = x and (g ∘ f)(x) = x, so function composition is commutative. Given f(x) = x² and g(x)=StartRoot x EndRoot, (f ∘ g)(0) = 0 and (g ∘ f)(0) = 0, so function composition is not commutative. Given f(x) = 4x and g(x) = x², (f ∘ g)(x) = 4x² and (g ∘ f)(x) = 16x², so function composition is not commutative.

The correct option are;

f(x) = x² - 4 and g(x) = x - 3, (f ο g)(2) = -3 and (g ο f)(2) = -3, so the function is commutative

Given f(x) = 4·x and g(x) = x², (f ο g)(x) = 4·x² and (g ο f)(x) = 16·x²

So the function is not commutative

Step-by-step explanation:

For the equations f(x) = x² - 4 and g(x) = x - 3, we have;

(f ο g)(x) = f(g(x)) = (x - 3)² - 4 = x² - 6·x + 9 - 4 = x² - 6·x + 5

At x = 2, we have;

(f ο g)(2) = f(g(2)) = 2² - 6×2 + 5 = - 3

Similarly, we have;

(g ο f)(x) = g(f(x)) = x² - 4 -3 = x² - 7

At x = 2, we have;

(g ο f)(2) = g(f(2)) = 2² - 7 = 4 - 7 = -3

Therefore, by commutative property, we have that the result of an operation does not change by changing the order of the operands such that we have;

a + b = b + a or a·b = b·a from which we have resolved also the following operation is commutative

(f ο g)(2) = (g ο f)(2)

Similarly given f(x) = 4·x and g(x) = x², (f ο g)(x) = 4·x² and (g ο f)(x) = 16·x²

So the function is not commutative

(f ο g)(x) = f(g(x)) = 4·x²

(f ο g)(x) = 4·x²

(g ο f)(x) = g(f(x)) = (4·x)² = 16·x²

(g ο f)(x) = 16·x²

∴ (f ο g)(x) = 4·x² ≠ (g ο f)(x) = 16·x²

(f ο g)(x)  ≠ (g ο f)(x) the function is not commutative.