Examine following function for maximum and minimum $f (x) = x^{3} - 9 x^{2} + 24 x$ .

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Solution

$f (x) = x^{3} - 9 x^{2} + 24 x$
$∴ f^{'} (x) = \frac{d}{d x} (x^{3} - 9 x^{2} + 24 x)$
$= 3 x^{2} - 9 \times 2 x + 24 \times 1$
$= 3 x^{2} - 18 x + 24$
and
$f^{''} (x) = \frac{d}{d x} (3 x^{2} - 18 x) + 24$
$= 3 \times 2 x - 18 \times 1 + 0$
$= 6 x - 18$
$f^{''} (x) = 0$ given $3 x^{2} - 18 x + 24 = 0$
$∴ x^{2} - 6 x + 8 = 0$
$∴ (x - 2) (x - 4) = 0$
$∴$ the roots of $f^{'} (x) = 0$ and $x_{1} = 2$ and $x_{2} = 4.$
(a) $f^{''} (2)$
$= 6 (2) - 18$
$= - 6 < 0$
$∴$ by the second derivatives test, $f$ has maximum at $x = 2$ and maximum value of at $x = 2$
$= f (2)$
$= (2)^{3} - 9 (2)^{2} + 24 (2)$
$= 8 - 36 + 48$
$= 20$
(b) $f^{''} (4) = 6 (4) - 18 = 6 > 0$
$∴$ by the second derivative test, $f$ has maximum at $x = 4$ and maximum value of at $x = 4$
$= f (4)$
$= (4)^{3} - 9 (4)^{2} + 24 (4)$
$= 64 - 144 + 96$
$= 16.$
Hence, the function $f$ has maximum value $20$ at $x = 2$ and minimum value $16$ at $x = 4.$

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