# Find all critical numbers of the following function: f(x) = 2x^3 + 3x^2 - 36x + 8 .

## Question:

Find all critical numbers of the following function: {eq}f(x) = 2x^3 + 3x^2 - 36x + 8 {/eq} .

## Critical Points:

It is at the critical points where the local extremes of the function are reached:

for differentiable functions these points are located only at the points where the derivative is equal to zero.

Calculating the derivative of the function:

{eq}f\left( x \right) = 2{x^3} + 3{x^2} - 36x + 8\\ \\ f'\left( x \right) = 6{x^2} + 6x - 36 {/eq}

And, equating to zero, we can calculate the critical points:

{eq}f'\left( x \right) = 6\left( {{x^2} + x - 6} \right) = 0\\ \\ {x^2} + x - 6 = 0\\ x = \frac{{ - 1 \pm \sqrt {{1^2} - 4 \cdot 1 \cdot \left( { - 6} \right)} }}{{2 \cdot 1}}\\ x = \frac{{ - 1 \pm \sqrt {25} }}{2} = \left\{ \begin{array}{l} 2\\ - 3 \end{array} \right. {/eq}