Question

F_1(x, y) = (y, x), f_2(x, y) = x + y and f_3(x, y) = x + 1.

(c) Does the function \( f_{2} \circ f_{1} \) have an inverse? Justify your answer carefully. (d) Assume \( g: \mathbb{R} \rightarrow \mathbb{R} \) is a function where \( g(x)=x^{2} \). Let \[ h:=g \c

Answer 1

The composition f_2 \circ f_1 is invertible because it is a bijective function; it switches and adds elements of an ordered pair, maintaining a one-to-one correspondence with unique inputs and outputs.

The original question seems to be about verifying the invertibility of a composite function f_2 \circ f_1 and working with another function g defined by g(x) = x^2. To determine whether f_2 \circ f_1 has an inverse, we need to check if it is a bijective function — both injective (one-to-one) and surjective (onto). However, the question seems incomplete as it does not specify the composition f_2 \circ f_1 explicitly and part (d) mentions h without further clarification.

Assuming standard compositions, f_1(x, y) = (y, x) simply switches elements, and f_2(x, y) = x + y adds elements of an ordered pair. The composition f_2(f_1(x, y)) would then be f_2(y, x) = y + x, which is essentially the same as f_2(x, y) due to commutative property of addition. This function is invertible because for every output there is a unique input pair making it bijective.