1 problem found
Let \[ \f(x) = 3ax^2 - 6x^3\, \] and, for each real number \(a\), let \({\rm M}(a)\) be the greatest value of \(\f(x)\) in the interval \(-\frac13 \le x \le 1\). Determine \({\rm M} (a)\) for \(a\ge0\). [The formula for \({\rm M} (a)\) is different in different ranges of \(a\); you will need to identify three ranges.]
Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}