## anonymous 5 years ago A(t)=t^5-5t^4+20t^3-17 I am having trouble setting the derivative equal to zero. Please help!

1. anonymous

What did you get for the derivative?

2. anonymous

I've got $A' = 5t^4-20t^3 + 60t^2 = 5t^2(t^2-4t + 12)$

3. anonymous

THat's exactly what I've got

4. myininaya

a*b=0 implies either a=0 or b=0 or both=0

5. anonymous

Well I know that 5t^2=0 but I when I solve the parentheses for 0 (using quadratic formula) it comes out negative. I can't fit that into the rest of my problem!

6. anonymous

(I am solving for the intervals of increasing and decreasing in the function)

7. myininaya

you mean the number under the square root is negative?

8. anonymous

Yes

9. myininaya

so it is imaginary that means your only critical number is 0

10. anonymous

That just means that that part of the equation doesn't have any real roots.

11. anonymous

Okay. So if I am solving for the intervals, do I just divide my number in half?

12. myininaya

I like using test points before and after each critical number

13. anonymous

You're looking for intervals for what? Where it's increasing and decreasing?

14. anonymous

Yes, @polpak

15. myininaya

the derivative tells us the slope therefore tells us if is increasing or decreasing so choose a number before and after 0 to see what if it is increasing or decreasing

16. anonymous

So it's increasing from negative infinity to infinity, correct?

17. anonymous

The derivative is always positive except at 0. So it is increasing for $$x \in (-\infty,0) \bigcup (0,\infty)$$

18. anonymous

It is not increasing at 0.

19. myininaya

it has a horizontal tangent at x=0 thats what we found when we set A'=0

20. myininaya

A'=0 means we are trying to find all x that has slope 0

21. anonymous

Okay, thanks guys!