# A function f : R

We have f : R R and f(x) = x3 + 4.

Recall that a function is a bijection only if it is both one-one and onto.

First, we will check if f is one-one.

Let x1, x2ϵ R (domain) such that f(x1) = f(x2)

x13 + 4 = x23 + 4

x13 = x23

(x1 – x2)(x12 + x1x2 + x22) = 0

As x1, x2ϵ R and the second factor has no real roots,

x1 – x2 = 0

x1 = x2

So, we have f(x1) = f(x2) x1 = x2.

Thus, function f is one-one.

Now, we will check if f is onto.

Let y ϵ R (co-domain) such that f(x) = y

x3 + 4 = y

x3 = y – 4 Clearly, for every y ϵ R, there exists x ϵ R (domain) such that f(x) = y and hence, function f is onto.

Thus, the function f is a bijection and has an inverse.

We have f(x) = y x = f-1(y)

But, we found f(x) = y Hence, Thus, f(x) is invertible and Hence, we have Thus, f-1(3) = –1.

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