Question

Is f(s)=x^3 a bijection from RR to RR?

Transformations of functions
ANSWERED
asked 2021-09-13
Is \(\displaystyle{f{{\left({s}\right)}}}={x}^{{3}}\) a bijection from \(\displaystyle\mathbb{R}\) to \(\displaystyle\mathbb{R}\)?

Expert Answers (1)

2021-09-14
\(\displaystyle{f}:{A}\to{B}\) is one-to-one if \(\displaystyle∀{a},{b}\in{A},{f{{\left({a}\right)}}}={f{{\left({b}\right)}}}\to{a}={b}\)
Assume that \(\displaystyle∀{x},{y}\in\mathbb{R},{f{{\left({x}\right)}}}={f{{\left({y}\right)}}}.\)
f(x)=f(y)
\(\displaystyle{x}^{{3}}={y}^{{3}}\)
x=y
Thus,
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is one-to-one
"For the function to be onto:
f(x)=y
let \(\displaystyle{y}\in\mathbb{R}\)
Choose x to be \(\displaystyle{y}^{{\frac{{1}}{{3}}}}.\)
Then,
\(\displaystyle{x}\in\mathbb{R}\)
and
\(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\)
\(\displaystyle={\left({y}^{{\frac{{1}}{{3}}}}\right)}^{{3}}\)
=y
Thus \(\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}\) is onto
Since it's both one-to-one and onto, it's a bijection function from \(\displaystyle\mathbb{R}\) to \(\displaystyle\mathbb{R}\)."
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