## mathcalculus Group Title find the critical points: s(t)= (t-1)^4 (t+5)^3 one year ago one year ago

1. mathcalculus Group Title

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

2. kobesaurus Group Title

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

3. kobesaurus Group Title

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

4. mathcalculus Group Title

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

5. kobesaurus Group Title

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

6. mathcalculus Group Title

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

7. kobesaurus Group Title

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

8. mathcalculus Group Title

yes

9. kobesaurus Group Title

kk

10. mathcalculus Group Title

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

11. kobesaurus Group Title

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

12. mathcalculus Group Title

how?

13. kobesaurus Group Title

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

14. Azteck Group Title

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. Azteck Group Title

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

16. Azteck Group Title

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. mathcalculus Group Title

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