# If f : R <span la

We have f : R R and f(x) = x3 – 3.

Recall that a function is invertible only when it is both one-one and onto.

First, we will prove that f is one-one.

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

x13 – 3 = x23 – 3

x13 = x23

(x1 – x2)(x12 + x1x2 + x22) = 0

x1 – x2 = 0 (as x1, x2ϵ R+)

x1 = x2

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

Thus, function f is one-one.

Now, we will prove that f is onto.

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

x3 – 3 = y

x3 = y + 3 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 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(24) = 3 and f-1(5) = 2.

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