Abram Boyd

2022-06-24

How do you algebraically demostrate that $f\left(x\right)={x}^{3}-x$ is not a one to one function?
So, I'm trying to prove that the function its not one to one, but I'm stuck:
$f\left({x}_{1}\right)=f\left({x}_{2}\right)\to {x}_{1}={x}_{2}$ then is a one to one
So ${x}_{1}^{3}-{x}_{1}={x}_{2}^{3}-{x}_{2}$
${x}_{1}^{3}-{x}_{2}^{3}-{x}_{1}+{x}_{2}=0$
$\left({x}_{1}-{x}_{2}\right)\left({x}_{1}^{2}+{x}_{1}{x}_{2}+{x}_{2}^{2}\right)-\left({x}_{1}-{x}_{2}\right)=0$
$\left({x}_{1}-{x}_{2}\right)\left({x}_{1}^{2}+{x}_{1}{x}_{2}+{x}_{2}^{2}-1\right)=0$
Which is true if ${x}_{1}={x}_{2}$ but how do I go about proving that is not a one to one function?

Nia Molina

Expert

Explanation:
You can assume f to be one to one and show contradictions.
such that ${0}^{3}-0=0$ and ${1}^{3}-1=0$ which contradicts f being a one to one function. Hence, f is not one to one.

fabios3

Expert

Explanation:
This is also true if ${x}_{1}^{2}+{x}_{1}\cdot {x}_{2}+{x}_{2}^{2}-1=0$
With the p-q formula, treating ${x}_{1}$ as the variable, we get
${x}_{1}=-\frac{{x}_{2}}{2}+\sqrt{\frac{{x}_{2}^{2}}{4}-{x}_{2}^{2}+1}=-\frac{{x}_{2}}{2}+\sqrt{1-\frac{3{x}_{2}^{2}}{4}}$
and ${x}_{1}=-\frac{{x}_{2}}{2}+\sqrt{1-\frac{3{x}_{2}^{2}}{4}}$
Thus, there are solutions when ${x}_{1}\ne {x}_{2}$ and you are not one to one.

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