1) For \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}\)

\(\displaystyle{F}{\left(-{1}\right)}={\left(-{1}\right)}^{{2}}={1}{\quad\text{and}\quad}{f{{\left({1}\right)}}}={1}^{{2}}={1}\)

\(\displaystyle{x}{1}\ne{x}{2},{b}{u}{t}{f{{\left({x}{1}\right)}}}={f{{\left({x}{2}\right)}}}\), so the function is not One-to-One, it doesn’t have an inverse.

2) For \(\displaystyle{g{{\left({x}\right)}}}={x}^{{3}}\)

\(\displaystyle{G}{\left(-{1}\right)}={\left(-{1}\right)}^{{3}}=-{1}{\quad\text{and}\quad}{g{{\left({1}\right)}}}={1}^{{3}}={1}\)

\(\displaystyle{X}{1}\ne{x}{2}{\quad\text{and}\quad}{g{{\left({x}{1}\right)}}}\ne{g{{\left({x}{2}\right)}}}\), so function is One-to-One, it has an inverse.

\(\displaystyle{F}{\left(-{1}\right)}={\left(-{1}\right)}^{{2}}={1}{\quad\text{and}\quad}{f{{\left({1}\right)}}}={1}^{{2}}={1}\)

\(\displaystyle{x}{1}\ne{x}{2},{b}{u}{t}{f{{\left({x}{1}\right)}}}={f{{\left({x}{2}\right)}}}\), so the function is not One-to-One, it doesn’t have an inverse.

2) For \(\displaystyle{g{{\left({x}\right)}}}={x}^{{3}}\)

\(\displaystyle{G}{\left(-{1}\right)}={\left(-{1}\right)}^{{3}}=-{1}{\quad\text{and}\quad}{g{{\left({1}\right)}}}={1}^{{3}}={1}\)

\(\displaystyle{X}{1}\ne{x}{2}{\quad\text{and}\quad}{g{{\left({x}{1}\right)}}}\ne{g{{\left({x}{2}\right)}}}\), so function is One-to-One, it has an inverse.