For what real values of $a$ and $b$ are all the extrema of the function $f (x) = a^{2} x^{3} - 0.5 a x^{2} - 2 x - b$ positive and the minimum is at the point $x_{0} = \frac{1}{3}$ ?

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Solution

$f^{'} (x) = 3 a^{2} x^{2} - a x - 2$
$f^{'} (1 / 3) = 0$
$\frac{3 a^{2}}{9} - \frac{a}{3} - 2 = 0 \frac{a^{2}}{9} - \frac{a}{3} = 2 a^{2} - a - 6 = 0 a^{2} - 3 a + 2 a - 6 = 0 (a - 3) (a + 2) = 0 a = 3 o r a = - 2$
when $a = 3$
$f (\frac{1}{3}) = \frac{9}{3^{3}} - \frac{3}{2 \times 9} - \frac{2}{3} - b > 0 b ϵ (- \infty, - 1 / 2)$
when $a = - 2$
Similarly $f (\frac{1}{3}) > 0 b ϵ (- \infty, - 11 / 27)$

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