Let R be the set of real numbers and f : R → R be the function defined by f (x) = 4x + 5.Show that f is invertible and find f −1

To show that a function is invertible, we need to show that it is both injective (one-to-one) and surjective (onto).

1. Injective: A function f: X → Y is said to be injective if for every x1, x2 in X, if f(x1) = f(x2), then x1 = x2. For our function f(x) = 4x + 5, let's assume f(x1) = f(x2). This gives us 4x1 + 5 = 4x2 + 5. Subtracting 5 from both sides, we get 4x1 = 4x2. Dividing by 4, we get x1 = x2. Hence, the function is injective.

2. Surjective: A function f: X → Y is said to be surjective if for every y in Y, there exists at least one x in X such that f(x) = y. For our function f(x) = 4x + 5, for any real number y, we can find an x = (y - 5) / 4 such that f(x) = y. Hence, the function is surjective.

Since the function is both injective and surjective, it is invertible.

Now, to find the inverse of the function f(x) = 4x + 5, we need to find a function f^(-1)(x) such that f(f^(-1)(x)) = x for all x in the domain of f, and f^(-1)(f(x)) = x for all x in the domain of f^(-1).

Let y = f(x) = 4x + 5. Solving for x, we get x = (y - 5) / 4. This gives us the inverse function f^(-1)(y) = (y - 5) / 4.