# Finding Critical numbers

1. Jan 5, 2005

### ziddy83

Hi,
i was wondering if anyone could help me.
The problem says find the critical numbers of f...
f(x) = 2x^3 + 15x^2 - 36x +1

ok i found f ' .....6x^2 + 30x-36,
Now how do i start on finding the critical numbers? Do i have to take the second derivative?

2. Jan 5, 2005

### Tide

What, exactly, is the definition of "critical number?" :-)

3. Jan 5, 2005

### da_willem

So should you take a second derivative?

4. Jan 5, 2005

### ziddy83

Right....its not, so I can just set it equal to zero, and then solve for the variable, right?

5. Jan 5, 2005

### dextercioby

Yes,your 'critical numbers' are solution of the equation
$$f'(x)=0$$
.BTW,your terminology is pretty weird.I use to call them "critical points",coz they pop up whenever i want to draw the graph of a function

Daniel.

6. Jan 5, 2005

### HallsofIvy

Staff Emeritus
"Right....its not, so I can just set it equal to zero, and then solve for the variable, right?"

Having determined first that there are no values of x for which the derivative does not exist, yes.

"BTW,your terminology is pretty weird.I use to call them "critical points",coz they pop up whenever i want to draw the graph of a function"

Actually, there is a difference. Given y= f(x), the "critical numbers" are the values of x at which f ' (x) does not exist or is equal to 0. The "critical points" are the points (x,y) with x a critical number.