Question

If $f:R\to R$ be the function defined by $f\left(x\right)=4x^3+7$, then show that $f$ is a bijection.

Answer 1

A function is bijective if it is both one-one and onto.

The function is $f:R\to R$ such that

$f\left(x\right)=4x^3+7$

Onto. Let $x_1,x_2\in A$

$f\left(x_1\right)=f\left(x_2\right)\forall x_1,x_2\in R$

$\Rightarrow 4x_1^3+7=4x_2^3+7$

$\Rightarrow 4x_1^3=4x_2^3\Rightarrow x_1^3-x_2^3=0$

$\Rightarrow (x_1-x_2)(x_1^2+x_1{x}_2+x_2^2)=0$ [Using $a^3-b^3=(a-b)(a^2+ab+b^2)$]

$\Rightarrow (x_1-x_2)\{{[x_1+\frac{x_2}{2}]}^2+\frac{3x_2^2}{4}\}=0$

$\Rightarrow x_1-x_2=0,$ because ${[x_1+\frac{x_2}{2}]}^2+\frac{3x_2^2}{4}\ne 0$

Since $f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$ ( as proved above ), therefore $f\left(x\right)$ is a one-one function.