Question

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

Answer 1

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

Answer 2

is it just \[(t-1)^4(t+5)^2\] ?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yes

Answer 9

kk

Answer 10

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

Answer 11

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

Answer 12

how?

Answer 13

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

Answer 14

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
\[\huge (t-1)^3(t+5)^2(7t+17)=0\]
\[\huge t=1, t=-5, t=-\frac{17}{7}\]

Answer 15

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

Answer 16

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.

Answer 17

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