What are the local extrema of f(x)= x^3 - 9x^2 + 19x - 3 ?

Feb 2, 2017

$f {\left(x\right)}_{\max} = \left(1.37 , 8.71\right)$
$f {\left(x\right)}_{\min} = \left(4.63 , - 8.71\right)$

Explanation:

$f \left(x\right) = {x}^{3} - 9 {x}^{2} + 19 x - 3$

$f ' \left(x\right) = 3 {x}^{2} - 18 x + 19$

$f ' ' \left(x\right) = 6 x - 18$

For local maxima or minima: $f ' \left(x\right) = 0$

Thus: $3 {x}^{2} - 18 x + 19 = 0$

Applying the quadratic formula:

$x = \frac{18 \pm \sqrt{{18}^{2} - 4 \times 3 \times 19}}{6}$

$x = \frac{18 \pm \sqrt{96}}{6}$

$x = 3 \pm \frac{2}{3} \sqrt{6}$

$x \cong 1.367 \mathmr{and} 4.633$

To test for local maximum or minimum:

$f ' ' \left(1.367\right) < 0 \to$ Local Maximum

$f ' ' \left(4.633\right) > 0 \to$ Local Minimum

$f \left(1.367\right) \cong 8.71$ Local Maximum
$f \left(4.633\right) \cong - 8.71$ Local Minimum

These local extrema can be seen on the graph of $f \left(x\right)$ below.

graph{ x^3-9x^2+19x-3 [-22.99, 22.65, -10.94, 11.87]}