Show that the function f:R→R given by f(x)=x3+x is a bijective function

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

 

  

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