Question

Let f₁ :R×R→R×R and f₂, f₃ :R×R→R be two functions where f₁(x,y)=(y,x),f₂(x,y)=x+y and f₃(x,y)=x+1. (a) Verify which functions f₁, f₂ and f₃ have inverse? If your answer is ‘fi has an inverse’, calculate the inverse. If your answer is ‘fi does not have an inverse’ justify your answer carefully. (b)Which composition is possible and which one is impossible: i. f₁◦ f₂; ii. f₂ ◦ f₃; iii. f₃ ◦ f₁. (c) Does the function f₂∘f₁ have an inverse? Justify your answer carefully. (d) Assume g : R→R is a function where g(x)=x2 . Let h : =g∘f ₂∘f₁∘f₁. i. Compute the value of h for (1,2) and (2,1), separately. ii. Compute the explicit formula for h((x,x)).

Answer 1

In this mathematics question, we determine which functions have inverses, verify possible compositions, and compute the value and explicit formula for a composition involving a given function.

Step-by-step explanation:

(a) To determine if a function has an inverse, we need to check if the function is one-to-one (injective) and onto (surjective).
f₁(x,y) = (y,x) is a one-to-one and onto function, so it has an inverse. The inverse of f₁ is f₁^(-1)(x,y) = (y,x).
f₂(x,y) = x+y is not a one-to-one function, so it does not have an inverse.
f₃(x,y) = x+1 is a one-to-one and onto function, so it has an inverse. The inverse of f₃ is f₃^(-1)(x,y) = (x-1,y).

(b) i. f₁◦f₂ is possible. The composition f₁◦f₂ is f₁(f₂(x,y)) = f₁(x+y,x) = (x,x+y).
ii. f₂◦f₃ is possible. The composition f₂◦f₃ is f₂(f₃(x,y)) = f₂(x+1,y) = x+1+y = x+y+1.
iii. f₃◦f₁ is impossible because f₁(x,y) = (y,x) and f₃(x,y) = x+1 do not have matching ranges.

(c) The composition f₂∘f₁ is not one-to-one, so it does not have an inverse. This can be seen from the fact that f₂∘f₁(x,y) = f₂(f₁(x,y)) = f₂(y,x) = y+x is not one-to-one.

(d) i. g(x) = x², f₂∘f₁(x,y) = f₂(f₁(x,y)) = f₂(y,x) = y+x, h = g∘f₂∘f₁∘f₁.
For (1,2): h(1,2) = g(f₂(f₁(f₁(1,2)))) = g(f₂(f₁(2,1))) = g(f₂(1,2)) = g(1+2,1) = g(3,1) = 3² = 9.
For (2,1): h(2,1) = g(f₂(f₁(f₁(2,1)))) = g(f₂(f₁(1,2))) = g(f₂(2,1)) = g(2+1,2) = g(3,2) = 3² = 9.

ii. The explicit formula for h((x,x)) = g(f₂(f₁(f₁(x,x)))) = g(f₂(f₁(x,x))) = g(f₂(x,x)) = g(x+x,x) = g(2x,x) = (2x)² = 4x².