Question

1. Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places.
2. Use calculus to find the exact maximum and minimum values.

71. \(f\left( x \right) = x\sqrt {x - {x^{\bf{2}}}} \)

Short answer

1. The point \(f\left( {0.75} \right) = 0.32\) is the abosolute maximum. The point \(f\left( 0 \right) = f\left( 1 \right) = 0\) is the absolute minimum.

2. The maximum value of \(f\left( x \right)\) is \(\frac{{3\sqrt 3 }}{{16}}\) at \(x = \frac{3}{4}\) and the minimum value is 0 at \(x = 0,{\rm{ and }}1\).

Step-by-step solution

Find the domain of \(f\left( x \right)\)

The function \(f\left( x \right)\) is defined, if;

\(\begin{array}{c}x - {x^2} \ge 0\\x\left( {1 - x} \right) \le 0\\0 \le x \le 1\end{array}\)

Sketch the graph of \(f\left( x \right)\)

In the desmos graphing calculator (for closed interval) enter the expression \(x\sqrt {x - {x^2}} \) in the tab and enter the interval \(0 \le x \le 1\), to plot the curve of \(f\left( x \right) = x\sqrt {x - {x^2}} \).

The figure below represents the graph of \(f\left( x \right)\).

Find the point of maximum and minimum of \(f\left( x \right)\)

Following observations can be made from the graph of \(f\left( x \right)\).

1. The point \(f\left( {0.75} \right) = 0.32\) is the abosolute maximum.
2. The point \(f\left( 0 \right) = f\left( 1 \right) = 0\) is the absolute minimum.

Differentiate the function \(f\left( x \right)\)

Find the derivative of \(f\left( x \right)\).

\(\begin{aligned}{c}f'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x\sqrt {x - {x^2}} } \right)\\ &= x\left( {\frac{1}{{2\sqrt {x - {x^2}} }}} \right)\left( {1 - 2x} \right) + \sqrt {x - {x^2}} \left( 1 \right)\\ &= \frac{{x\left( {1 - 2x} \right) + 2x - 2{x^2}}}{{2\sqrt {x - {x^2}} }}\\ &= \frac{{3x - 4{x^2}}}{{2\sqrt {x - {x^2}} }}\end{aligned}\)

Find the root of the equation \(f'\left( x \right) = 0\).

\(\begin{aligned}{c}3x - 4{x^2} &= 0\\x\left( {3 - 4x} \right) &= 0\\x &= 0,\frac{3}{4}\end{aligned}\)

Find the values of function at critical points of \(f\left( x \right)\)

The value of \(f\left( x \right)\) at \(x = 1\);

\(\begin{aligned}{c}f\left( 1 \right) &= 1\sqrt {1 - 1} \\ &= 0\end{aligned}\)

At \(x = 0\),

\(\begin{aligned}{c}f\left( 0 \right) &= 0\sqrt {0 - {0^2}} \\ &= 0\end{aligned}\)

At \(x = \frac{3}{4}\),

\(\begin{aligned}{c}f\left( {\frac{3}{4}} \right) &= \frac{3}{4}\sqrt {\frac{3}{4} - {{\left( {\frac{3}{4}} \right)}^2}} \\ &= \frac{3}{4}\sqrt {\frac{3}{4} - \frac{9}{{16}}} \\ &= \frac{3}{4}\sqrt {\frac{3}{{16}}} \\ &= \frac{{3\sqrt 3 }}{{16}}\end{aligned}\)

Thus, the maximum value of \(f\left( x \right)\) is \(\frac{{3\sqrt 3 }}{{16}}\) at \(x = \frac{3}{4}\) and the minimum value is 0 at \(x = 0,{\rm{ and }}1\).