# How do you find the critical points f(x)= x^3 + 35x^2 - 125x - 9375?

Jun 27, 2015

The critical numbers are: $x = - 25$ and $x = \frac{5}{3}$

#### Explanation:

A critical number for $f$ is a number in the domain of $f$ at which, $f ' = 0$ or $f '$ does not exist.

$f \left(x\right) = {x}^{3} + 35 {x}^{2} - 125 x - 9375$

$f ' \left(x\right) = 3 {x}^{2} + 70 x - 125$

Clearly, this function exists for all $x$, so we need only consider its zeros:

$3 {x}^{2} + 70 x - 125 = 0$

To look for factors using whole numbers:
$3 \times - 125 = - 375$
Find two numbers whose product is $- 375$ and whose sum is $70$

$- 1 \times 375$ won't work
$- 2$ is not a factor of 375
$- 3 \times 125$ won't work
$- 5 \times 75$ STOP! that's the one.
Now split the middle term and factor by grouping:

$3 {x}^{2} + 70 x - 125 = 0$

$3 {x}^{2} - 5 x + 75 x - 125 = 0$

$x \left(3 x - 5\right) + 25 \left(3 x - 5\right) = 0$

$\left(x + 25\right) \left(3 x - 5\right) = 0$

$x = - 25$ and $x = \frac{5}{3}$. Both are in the domain of $f$ (Domain of $f = \mathbb{R}$), so they are the critical numbers for $f$.