What is the minimum vertical distance between the parabolas y=x^2+1 and PS

tabita57i 2021-09-22 Answered
What is the minimum vertical distance between the parabolas \(\displaystyle{y}={x}^{{2}}+{1}\) and \(\displaystyle{y}={x}-{x}^{{2}}\)

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Expert Answer

Sally Cresswell
Answered 2021-09-23 Author has 4161 answers
\(\displaystyle{y}_{{1}}={x}^{{1}}+{1},\ {y}_{{2}}={x}-{x}^{{2}}\)
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: \(\displaystyle{\frac{{{7}}}{{{8}}}}\)
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