Question

If f:R where R is te function defined by f(x)=4x^3 +7,then show that f is bijection​

Answer 1

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

The function is f:R→R such that

$\bold{f(x)=4x^3+7}$

$\bold{Onto. \: Let \: x_1 ,x_2∈A}$

$\bold{f(x_1)=f(x_2)∀x_1 ,x_2 ∈R}$

$\bold{⇒4x^3_1+7=4x^3_2+7}$

$\bold{⇒4x^3_1 = 4x^3_2⇒x^3_1 - x^3_2 = 0}$

$\small \bold{⇒(x_1 −x_2)(x^2_1+x_1x_2+x^2_2)=0}$

$\small\bold{⇒(x_1−x_2) \{[x_1+\frac{x_2}{2}]^2+\frac{3x^2_2}{4} \}\ne0}$

$\small \bold{Since \: f(x_1)=f(x_2)⇒x_1=x_2 ( as \: proved}$

above ), therefore f(x) is a one-one function.