Question

What is the minimum vertical distance between the parabolas \(y = {x^2} + 1\) and \(y = x - {x^2}\)?

Short answer

The minimum distance is \(\frac{7}{8}{\rm{ units}}\).

Step-by-step solution

Optimization of a Function

The Optimization of a Function \(f\left( x \right)\) for any real value of \(c\) can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\).

Optimizing the function according to the given condition.

The given data can be examined with the help of graph as:

Let the vertical distance here be:

\(\begin{aligned}{c}D\left( x \right) = \left( {{x^2} + 1} \right) - \left( {x - {x^2}} \right)\\ = 2{x^2} - x + 1\end{aligned}\).

For minimization, we have:

\(\begin{aligned}{c}D'\left( x \right) = \frac{d}{{dx}}\left( {2{x^2} - x + 1} \right)\\\frac{d}{{dx}}\left( {2{x^2} - x + 1} \right) = 0\\4x - 1 = 0\\x = \frac{1}{4}\end{aligned}\)

Now, we have:

\(\begin{aligned}{l}{\rm{for }}x < \frac{1}{4} \to D'\left( x \right) < 0\\{\rm{for }}x > \frac{1}{4} \to D'\left( x \right) > 0\end{aligned}\)

Therefore, the function will have an absolute minimum value at \(x = \frac{1}{4}\).

And the minimum distance will be as:

\(\begin{aligned}{c}D\left( {\frac{1}{4}} \right) = 2{\left( {\frac{1}{4}} \right)^2} - \left( {\frac{1}{4}} \right) + 1\\ = \frac{1}{8} - \frac{1}{4} + 1\\ = \frac{7}{8}\end{aligned}\)

Hence, the minimum distance is \(\frac{7}{8}{\rm{ units}}\).