Question

show that the function f: R ---->R given by f(x)=x3 + x is a bijection.

Answer 1

.I'm fairly happy with what I've done (I think):

f:R→R, f(x)=x3

So we know that to prove if a function is bijective, we must prove it is both injective and surjective.

Proof: f is injective

Let:
x,y∈R : f(x) = f(y)

x3=y3

(take cube root of both sides)
x=y

Proof: f is surjective

Let:
y∈R

x=y√3

f(x)=(y√3)3=y

So I believe that is enough to prove bijectivity for f(x)=x3. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.


I hope it will help you..

Answer 2

Answer:

f:R->R f(x)=x3+x

Let x,y belongs to R

f(x)=f(y)

x3+x=y3+y

(x3-y3)+x-y=0

(x-y)(x2+xy+y2+1)=0

x=y

f(x)=f(y)

=) x=y for all x,y belongs to R

So, f is one-one function

Let y be any arbitrary element of R

f(x)=y

=) x3+x=y

x3+x-y=0

For every value of y, the equation x3+x-y=0 has a real root a (alpha)

a3+a-y=0

a3+a=y

f(a)=y

For every y belongs to R there exists a (alpha) belongs R such that f(a)=y

So, f is a onto function

Hence, f:R->R is a bijective function