Question

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

Answer 1

$f: \real \to \real$ 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