Q. 115.0( 3 Votes )
A function f : R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f-1(3).
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 ⇒
Thus, f(x) is invertible and
Hence, we have
Thus, f-1(3) = –1.
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