# Math Help - Critical Numbers

1. ## Critical Numbers

F(x)=2x^3 - 3x^2 +36x +7

I need to find all critical numbers. Thank you

2. Originally Posted by dimanre

F(x)=2x^3 - 3x^2 +36x +7

I need to find all critical numbers. Thank you

Set $f'(x)=0$

The values of x that satisfy this equation are your "critical #'s"

3. ## ok

so then I take 2(0)^3 - 3(0)^2 +36(0) +7

so that gives me 7? since all others become zero? or am i way off?

4. Originally Posted by dimanre
so then I take 2(0)^3 - 3(0)^2 +36(0) +7

so that gives me 7? since all others become zero? or am i way off?
You have to take the derivative first, then set it to zero, and then solve for x. That will you give your critical numbers.

5. Originally Posted by dimanre
so then I take 2(0)^3 - 3(0)^2 +36(0) +7

so that gives me 7? since all others become zero? or am i way off?
$f'(x)=6x^2-6x+36$

Do you know how I did this?

6. ## ok

ok so it becomes 6x^2 - 6x +36=0

is this correct? if so..what is next step?

7. Originally Posted by dimanre
ok so it becomes 6x^2 - 6x +36=0

is this correct? if so..what is next step?
Solve for x.

8. Ok..i figured out how to get the derivative...and then i know you have to set x to zero..so do you multiply each by zero? thereby giving you zero for each section?

9. Originally Posted by dimanre
Ok..i figured out how to get the derivative...and then i know you have to set x to zero..so do you multiply each by zero? thereby giving you zero for each section?
Not quite.

Set $f'(x)=0$ and then solve for x. At this point it's algebra man.

10. ## yes...which is very weak....

I know...but my algebra is horribly bad and rusty..havent taken it in 18 years....so i am stuck trying to figure it out...

11. I believe it is 6(x^2-x+6) is that the final answer?

12. Originally Posted by dimanre
I know...but my algebra is horribly bad and rusty..havent taken it in 18 years....so i am stuck trying to figure it out...
$6x^2-6x+36=0$

$x^2-x+6=0$

$x=\frac{1\pm\sqrt{1-4(1)(6)}}{2}$ (by the quadratic formula).

You can see that this has no solution, therefore the are no critical numbers.

13. You can clearly see from the graph that f(x) does not have a place where a tangent line will be horizontal.

14. THank you so much for your help..i really appreciate it

15. Originally Posted by dimanre
THank you so much for your help..i really appreciate it