Prove that f:R→ R defined by f(x)=x3 -5 is a bijection

$f:\Re \to \Re$ is a bijection if and only if f(x) is an injection and is Surjective.

Injection

If $f(x) = f(y) \implies x =y$ then f is an injection.

$x^3-5=y^3-5$$x^3=y^3$$x=y$

Therefore, f is an injection.

Surjection:

$f: A \to B$ is Surjective if for all $y \in B \exists x \in A: F(x) =y$

$y=x^3-5$$x={(y+5)}^{1 \over 3}$

$f(x) ={({(y+5)^{1 \over 3}})^3}-5$$=y+5-5$$=y$

Therefore f is Surjective

Thus, since $f: \real \to \real$ is both injective and Surjective, it is a bijection