# Finding stationary points for derivative tests

• Feb 24th 2009, 03:03 PM
mezhopking
Finding stationary points for derivative tests
Hi,

I'm having trouble working out the correct signs to give to stationary points (I can't see a pattern from examples).

I've derived and factored the function:

f(x) = -2x^3 - 21x^2 - 36x + 33

to

f'(x) = -6(x^2 + 7x + 6)

From here the equation f'(x) = 0 is equivalent to:
x^2 + 7x + 6 = 0; that is (x + 1)(x + 6) = 0

Are my stationary points x = 1 and x = 6 OR x = -1 and x = -6 ??

Obviously this is fundamental in achieving the correct answers for the First and Second Derivative tests...

From examples in the text I'm following, the signs in the brackets flip / invert... which could be seen as a first bracket / second bracket substitution OR just a general invert, as they don't provide an example where both the signs are the same (as in my example above).

Is the rule simply to invert, or is there something more to it?

Thanks you so much for taking the time to help me with this problem.

Mez
• Feb 24th 2009, 05:02 PM
arpitagarwal82
$(x +1) (x + 6) = 0$
means either $x + 1 =0$ or $x + 6 =0$
so $x + 1 = 0$ gives $x = -1$and $x + 6 = 0$ gives $x = -6$

Is that clear?
• Feb 25th 2009, 04:46 AM
mezhopking
Quote:

Originally Posted by arpitagarwal82
Is that clear?

Perfectly clear, and clearly obvious... thanks a lot for the help.