Invertible Functions and Finding the Inverse

1. Note domain and codomain. 2. Prove one-one using f(a) = f(b), or give a counterexample if disproving. 3. Prove onto by solving y = f(x) for x in the domain. 4. Once bijective, solve y = f(x) for x. 5. Write the inverse with correct domain and codomain. 6. Optionally verify using f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.

A reverse mapping diagram may support the idea of inverse, but the exam-useful test is bijectivity plus composition identity.

Forgetting that f⁻¹ has domain equal to the codomain of f, causing an inverse formula to be written with the wrong domain.

A function f: A to B is invertible if there exists a function f⁻¹: B to A such that f⁻¹(f(x)) = x for every x in A and f(f⁻¹(y)) = y for every y in B. A function is invertible if and only if it is bijective.

A function f: A to B is invertible if there exists a function f⁻¹: B to A such that f⁻¹(f(x)) = x for every x in A and f(f⁻¹(y)) = y for every y in B. A function is invertible if and only if it is bijective. Invertibility theorem: f is invertible if and only if f is bijective. Inverse identities: f⁻¹∘f = identity on A and f∘f⁻¹ = identity on B. Method: write y = f(x), solve for x in terms of y, then replace y by x after stating the inverse rule and domain. For f: R to R, f(x) = 5x - 7, set y = 5x - 7. Then x = (y + 7)/5. Hence f⁻¹(y) = (y + 7)/5, or f⁻¹(x) = (x + 7)/5.

An inverse function reverses the action of the original function. It exists only when every output in the codomain comes from exactly one input. One-one gives uniqueness, and onto gives existence. Both are necessary. Invertibility theorem: f is invertible if and only if f is bijective. Inverse identities: f⁻¹∘f = identity on A and f∘f⁻¹ = identity on B. Method: write y = f(x), solve for x in terms of y, then replace y by x after stating the inverse rule and domain. Let f: R to R be f(x) = 2x + 5. First prove bijective. One-one: f(a) = f(b) gives 2a + 5 = 2b + 5, so a = b. Onto: for any y in R, y = 2x + 5 gives x = (y - 5)/2, which belongs to R. Hence f is bijective and invertible. To find inverse, y = 2x + 5, so x = (y - 5)/2. Therefore f⁻¹(y) = (y - 5)/2, and f⁻¹(x) = (x - 5)/2. Asked as proof of invertibility, finding the inverse of a given function, or verifying that two functions are inverses using composition. Assertion-reason questions often test the statement that every bijection has an inverse. Forgetting that f⁻¹ has domain equal to the codomain of f, causing an inverse formula to be written with the wrong domain.