\(\displaystyle{y}_{{1}}={x}^{{1}}+{1},\ {y}_{{2}}={x}-{x}^{{2}}\)

The vertical at any point x is obtained form

\(\displaystyle{D}={y}_{{1}}-{y}_{{2}}={x}^{{2}}+{1}-{\left({x}-{x}^{{2}}\right)}\)

\(\displaystyle={x}^{{2}}+{1}-{x}+{x}^{{2}}\)

\(\displaystyle={2}{x}^{{2}}-{x}+{1}\)

If you subtract the other way the distance you get is negative because of where the graphs are situated, but the problem would work out similarly.

Find where \(\displaystyle{D}'={0}\)

\(\displaystyle{D}'{\left({x}\right)}={0}={4}{x}-{1}\)

\(\displaystyle{x}={\frac{{{1}}}{{{4}}}}\)

\(\displaystyle{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{1}\) is an upward opening parabola so \(\displaystyle{x}={\frac{{{1}}}{{{4}}}}\) is the minimum.

Plug in to get the minimum distance

\(\displaystyle{D}{\left({\frac{{{1}}}{{{4}}}}\right)}={2}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{2}}-{\frac{{{1}}}{{{4}}}}+{1}\)

\(\displaystyle={\frac{{{1}}}{{{8}}}}-{\frac{{{1}}}{{{4}}}}+{1}\)

\(\displaystyle={\frac{{{1}-{2}+{8}}}{{{8}}}}\)

\(\displaystyle={\frac{{{7}}}{{{8}}}}\)

Result: \(\displaystyle{\frac{{{7}}}{{{8}}}}\)

The vertical at any point x is obtained form

\(\displaystyle{D}={y}_{{1}}-{y}_{{2}}={x}^{{2}}+{1}-{\left({x}-{x}^{{2}}\right)}\)

\(\displaystyle={x}^{{2}}+{1}-{x}+{x}^{{2}}\)

\(\displaystyle={2}{x}^{{2}}-{x}+{1}\)

If you subtract the other way the distance you get is negative because of where the graphs are situated, but the problem would work out similarly.

Find where \(\displaystyle{D}'={0}\)

\(\displaystyle{D}'{\left({x}\right)}={0}={4}{x}-{1}\)

\(\displaystyle{x}={\frac{{{1}}}{{{4}}}}\)

\(\displaystyle{D}{\left({x}\right)}={2}{x}^{{2}}-{x}+{1}\) is an upward opening parabola so \(\displaystyle{x}={\frac{{{1}}}{{{4}}}}\) is the minimum.

Plug in to get the minimum distance

\(\displaystyle{D}{\left({\frac{{{1}}}{{{4}}}}\right)}={2}{\left({\frac{{{1}}}{{{4}}}}\right)}^{{2}}-{\frac{{{1}}}{{{4}}}}+{1}\)

\(\displaystyle={\frac{{{1}}}{{{8}}}}-{\frac{{{1}}}{{{4}}}}+{1}\)

\(\displaystyle={\frac{{{1}-{2}+{8}}}{{{8}}}}\)

\(\displaystyle={\frac{{{7}}}{{{8}}}}\)

Result: \(\displaystyle{\frac{{{7}}}{{{8}}}}\)