## anonymous 3 years ago find the critical points: s(t)= (t-1)^4 (t+5)^3

1. anonymous

i found the derivative. 4(t-1)^3 3(t+5)^2

2. anonymous

is it just $(t-1)^4(t+5)^2$ ?

3. anonymous

with the chain rule those are not the derivatives...

4. anonymous

yes but instead of 2 as an exponent its 3. i know im suppose to = 0 then im not sure what to do.

5. anonymous

wait that is right, t is single power, sorry.

6. anonymous

okay so i did it wrong can we review step by step

7. anonymous

just to be clear, it isnt the two equations plus one another, its them two multiplied by one another?

8. anonymous

yes

9. anonymous

kk

10. anonymous

so i get.... 4(t-1)^3 3(t+5)^2

11. anonymous

(t-1)^3 (t+5)^2 (17+7 t)

12. anonymous

how?

13. anonymous

I'm trying to figure that out, I used Wolframalpha to check my work and it got something different.

14. anonymous

Use the product rule!!! $\huge s(t)= (t-1)^4 (t+5)^3$ $\huge s'(t)=4(t-1)^3(t+5)^3+3(t+5)^2(t-1)^4$ $\huge =(t-1)^3(t+5)^2[4t+20+3t-3]$ $\huge =(t-1)^3(t+5)^2(7t+17)$ At S.P's s'(t)=0 $\huge (t-1)^3(t+5)^2(7t+17)=0$ $\huge t=1, t=-5, t=-\frac{17}{7}$

15. anonymous

$s(t)=y$ To find the values of y, sub in the values of t into the original equation.

16. anonymous

Don't think you need to describe the nature of the curve [ie. Maximum or minimum turning point, etc.]. But I think you still need to find the inflection points of the function.

17. anonymous

how did you get to this step: =(t−1)3(t+5)2[4t+20+3t−3]