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Show that the function f:R→R given by f(x)=x^{3}+x is a bijective function

^{3}+x

Let x,y belongs to R

f(x)=f(y)

x^{3}+x=y^{3}+y

(x^{3}-y^{3})+x-y=0

(x-y)(x^{2}+xy+y^{2}+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

=) x^{3}+x=y

x^{3}+x-y=0

For every value of y, the equation x^{3}+x-y=0 has a real root a (alpha)

a^{3}+a-y=0

a^{3}+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

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