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

1. mathcalculus

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

2. kobesaurus

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

3. kobesaurus

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

4. mathcalculus

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

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

6. mathcalculus

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

7. kobesaurus

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

8. mathcalculus

yes

9. kobesaurus

kk

10. mathcalculus

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

11. kobesaurus

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

12. mathcalculus

how?

13. kobesaurus

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

14. Azteck

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

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

16. Azteck

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

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