# How do you find the maximum, minimum and inflection points and concavity for the function f(x)=18x^3+5x^2-12x-17?

Jul 17, 2018

Below

#### Explanation:

$f \left(x\right) = 18 {x}^{3} + 5 {x}^{2} - 12 x - 17$

$f ' \left(x\right) = 54 {x}^{2} + 10 x - 12$

$f ' ' \left(x\right) = 108 x + 10$

For maximum or minimum points, $f ' \left(x\right) = 0$

$54 {x}^{2} + 10 x - 12 = 0$
$27 {x}^{2} + 5 x - 6 = 0$
$x = \frac{- 5 \pm \sqrt{25 + 648}}{54}$
$x = \frac{- 5 \pm \sqrt{673}}{54}$
$x = \frac{- 5 + \sqrt{673}}{54}$ or $x = \frac{- 5 - \sqrt{673}}{54}$

To determine whether the point is maximum or minimum,
At $\left(\frac{- 5 + \sqrt{673}}{54} , - 19.85\right)$,
$f ' ' \left(x\right) = 51.88 > 0$
Therefore, it is minimum and concave up at $\left(\frac{- 5 + \sqrt{673}}{54} , - 19.85\right)$

At $\left(\frac{- 5 - \sqrt{673}}{54} , - 0.57\right)$
$f ' ' \left(x\right) = - 51.88 < 0$
Therefore, it is maximum and concave down at $\left(\frac{- 5 - \sqrt{673}}{54} , - 0.57\right)$

For inflexion points, $f ' ' \left(x\right) = 0$

$108 x + 10 = 0$
$x = - \frac{10}{108} = - \frac{5}{54}$

Test $\left(- \frac{5}{54} , - 15.86\right)$
$f ' ' \left(0\right) = 0 + 10 = 10 > 0$
$f ' ' \left(- \frac{5}{54}\right) = 0$
$f ' ' \left(- \frac{1}{2}\right) = - 44 < 0$
Therefore, since there is a change in concavity, a point of inflexion occurs at $\left(- \frac{5}{54} , - 15.86\right)$