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

one-one function

$f(x)=x^3+x+1$

Let $f(x_1) = f(x_2)$

$⇒ x_1^3 + x_1 + 1 = x_2^3 + x_2 + 1$

$⇒x_1^3-x_2^3+x_1-x_2=0$

$⇒(x_1 − x_2) (x_1^2 + x_1 x_2+x_2^2+1)=0$

$⇒x_1= x_2$ or $x_1^2+x_1x_2+x_2^2+1=0$

From $x_1^2+x_1x_2+x_2^2+1=0$

$x_1=\frac{-x_2±\sqrt{x_2^2-4(x_2^2+1)}}{2}$

$=\frac{-x_2±\sqrt{-3x_2^2-4}}{2}$

$=\frac{-x_2±i\sqrt{3x_2^2+4}}{2}$, where $i=\sqrt{-1}$

Thus, $x_1 = x_2$ only.

Hence f(x) is one-one.

Also, $f'(x)=3x^2+1>0∀∈R$

So, f(x) is one-one function.