# How do you find the critical points f(x)= 2x^3 + 3x^2 - 36x + 5?

Aug 10, 2015

$\left(- 3 , 86\right)$ and $\left(2 , - 39\right)$

#### Explanation:

From Wikipedia: a critical point or stationary point of a differentiable function of a single real variable, $f \left(x\right)$, is a value ${x}_{0}$ in the domain of $f$ where its derivative is 0: f′(x_0) = 0

Thus, to find the critical points of $f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} - 36 x + 5$, we first need to compute $f ' \left(x\right)$ then find all the $x$-values such that $f ' \left(x\right) = 0$.

$f \left(x\right) = 2 {x}^{3} + 3 {x}^{2} - 36 x + 5$
$f ' \left(x\right) = 6 {x}^{2} + 6 x - 36 x$

When $f ' \left(x\right) = 0$:
$6 {x}^{2} + 6 x - 36 x = 0$
${x}^{2} + x - 6 x = 0$
$\left(x + 3\right) \left(x - 2\right) = 0$
$x = - 3$ or $x = 2$

$x = - 3$, $y = 2 {\left(- 3\right)}^{3} + 3 {\left(- 3\right)}^{2} - 36 \left(- 3\right) + 5 = 86$
$x = 2$, $y = 2 {\left(2\right)}^{3} + 3 {\left(2\right)}^{2} - 36 \left(2\right) + 5 = - 39$

Thus, critical points are $\left(- 3 , 86\right)$ and $\left(2 , - 39\right)$ You could conjecture that the first is a maximum point and the second a minimum, but you'll need the second derivative test to prove that in totality.