First take note of the domain of <em>f(x)</em> ; the square root term is defined as long as <em>x</em> - <em>x</em> ² ≥ 0, or 0 ≤ <em>x</em> ≤ 1.
Check the value of <em>f(x)</em> at these endpoints:
<em>f</em> (0) = 0
<em>f</em> (1) = 0
Take the derivative of <em>f(x)</em> :
For <em>x</em> ≠ 0, we can eliminate the √<em>x</em> term in the denominator:
<em>f(x)</em> has critical points where <em>f '(x)</em> is zero or undefined. We know about the undefined case, which occurs at the boundary of the domain of <em>f(x)</em>. Check where <em>f '(x)</em> = 0 :
√<em>x</em> (3 - 4<em>x</em>) = 0
√<em>x</em> = 0 <u>or</u> 3 - 4<em>x</em> = 0
The first case gives <em>x</em> = 0, which we ignore. The second leaves us with <em>x</em> = 3/4, at which point we get a maximum of max{<em>f(x) </em>} = 3√3 / 2.