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Hey guys could someone check my inverse question? would be highly appreciated, if theres anything i could do to improve the layout would be great too many thanks.
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p.s sorry about the weird format
You want to show that the function $f:[0,1] \to [0,2]$ given by $f(x)=x^4+x^2$ is bijective (and hece invertible.)
Injectivity: Let $a,b \in [0,1]$ such that $$f(a)=f(b) \Leftrightarrow a^2+a^4=b^2+b^4,$$ then $$f(a)=f(b) \Leftrightarrow (a^2-b^2)+(a^4-b^4)=0 \Leftrightarrow (a^2-b^2)+(a^2-b^2)(a^2+b^2)=0 \Leftrightarrow (a^2-b^2)(1+a^2+b^2)=0 \Leftrightarrow (a+b)(a-b)(a^2+b^2+1)=0.$$ Notice that the only posible solutions are $a=\pm b$ since $a^2+b^2+1 \neq 0$ for all $a,b \in \mathbb{R}$. But $a=-b$ is not possible because if $a \in [0,1]$ then $b \in [-1,0]$, a contradiciton. We have proved that $f(a)=f(b)$ iff $a=b$. Hence $f$ is injective.
Surjectivity: Let $y \in [0,2]$, we want to show that there exists $x \in [0,1]$ such that $f(x)=y$ i.e. $y=x^2+x^4$. Observe that $$0\leq x \leq 1\Leftrightarrow (0^2=0 \leq x^2 \leq 1^2=1) \: \wedge \: (0^4=0 \leq x^4 \leq 1^4=1) ,$$ therefore, adding both inequalities yields $$x \in [0,1] \Leftrightarrow 0 \leq x \leq 1 \Leftrightarrow 0 \leq x^2+x^4 \leq 2\Leftrightarrow 0 \leq y \leq 2 \Leftrightarrow y \in [0,2].$$ We conclude that $f$ is surjective.
Since $f$ is both one-to-one and onto, we conclude that $f$ is invertible.