# How do you find the exact relative maximum and minimum of the polynomial function of f(x) = x^3 + 4x^2 - 5x?

Dec 25, 2017

minimum: $\approx \left(0.523 , - 1.378\right)$

maximum: $\approx \left(- 3.189 , 24.193\right)$

#### Explanation:

relative max/min can only exist if $f ' \left(x\right)$ is 0 at that x-value.

first find $f ' \left(x\right)$:
$= 3 {x}^{2} + 8 x - 5$ (power rule)

solving this quadratic: $x = \frac{- 8 \pm \sqrt{{8}^{2} - 4 \left(3\right) \left(- 5\right)}}{2 \cdot 3}$
$x = \frac{- 8 \pm \sqrt{124}}{6}$
$x = \frac{- 8 \pm 2 \sqrt{31}}{6}$
$x = \frac{- 4 \pm \sqrt{31}}{3}$

since $f \left(x\right)$ is a cubic with a positive leading coefficient, it will approach $- \infty$ on the left and $+ \infty$ on the right. this means the lesser x-value where $f ' \left(x\right) = 0$ , ($\frac{- 4 - \sqrt{31}}{3}$) , is a local maximum and the greater x-value , ($\frac{- 4 + \sqrt{31}}{3}$) , is a local minimum.

plugging the x-values into $f \left(x\right)$ gives:
minimum: $\left(- \frac{4}{3} + \frac{\sqrt{31}}{3} , \frac{308}{27} - \frac{62 \sqrt{31}}{27}\right)$ $\approx \left(0.523 , - 1.378\right)$

maximum: $\left(- \frac{4}{3} - \frac{\sqrt{31}}{3} , \frac{308}{27} + \frac{62 \sqrt{31}}{27}\right)$ $\approx \left(- 3.189 , 24.193\right)$