Question
Classify the following functions as injection, surjection or bijection:f: $Q \rightarrow Q$, defined by $f(x)=x^3+1$
✓
Answer
$f: Q \rightarrow Q$, defined by $f(x)=x^3+1$ Injective: Let $x, y \in Q$ such that
$f(x) = f(y)$
$\Rightarrow x^3 + 1 = y^3 + 1$
$\Rightarrow (x^3 - y^3) = 0$
$\Rightarrow (x - y)(x^2 + xy + y^2) = 0$
but $\text{x}^2+\text{xy}+\text{y}^2\geq0$
$\therefore$ $x - y = 0$
$\Rightarrow x = y$
$\therefore$ f is injective.
Surjective: Let $\text{y}\in\text{Q}$ be arbitrary, then
$f(x) = y$
$\Rightarrow x^3 + 1 - y = 0$
We know that a degree 3 equation has alteast one real solution.
Let $\text{x}=\alpha$ be that solution
$\therefore\ \alpha^3+1=\text{y}$
$\therefore\ \text{f}(\alpha)=\text{y}$
$\therefore$ f is onto.
$f(x) = f(y)$
$\Rightarrow x^3 + 1 = y^3 + 1$
$\Rightarrow (x^3 - y^3) = 0$
$\Rightarrow (x - y)(x^2 + xy + y^2) = 0$
but $\text{x}^2+\text{xy}+\text{y}^2\geq0$
$\therefore$ $x - y = 0$
$\Rightarrow x = y$
$\therefore$ f is injective.
Surjective: Let $\text{y}\in\text{Q}$ be arbitrary, then
$f(x) = y$
$\Rightarrow x^3 + 1 - y = 0$
We know that a degree 3 equation has alteast one real solution.
Let $\text{x}=\alpha$ be that solution
$\therefore\ \alpha^3+1=\text{y}$
$\therefore\ \text{f}(\alpha)=\text{y}$
$\therefore$ f is onto.
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