Check the nature of the function ` f(x)=x^(3)+x+1, x in R ` using analytical method and differentiation method.

Explanation

`f(x)=x^(3)+x+1`
Let `f(x_(1))=f(x_(2))`
`impliesx_(1)^(3)+x_(1)+1=x_(2)^(3)+x_(2)+1`
`impliesx_(1)^(3)-x_(2)^(3)+x_(1)-x_(2)=0`
`implies (x_(1)-x_(2))(x_(1)^(2)+x_(1)x_(2)+x_(2)^(2)+1)=0`
`implies x_(1)=x_(2) " or " x_(1)^(2)+x_(1)x_(2)+x_(2)^(2)+1=0`
from `x_(1)^(2)+x_(1)x_(2)+x_(2)^(2)+1=0`
`x_(1)=(-x_(2)+-sqrt(x_(2)^(2)-4(x_(2)^(2)+1)))/(2)`
`=(-x_(2)+-sqrt(-3x_(2)^(2)-4))/(2)`
`=(-x_(2)+- isqrt(3x_(2)^(2)+4))/(2), " where " i=sqrt(-1)`
Thus, `x_(1)=x_(2)` only.
Hence f(x) is one-one.
Also, `f'(x)=3x^(2) +1 gt0 AA in R`.
So, f(x) is one-one function.