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

Calculus1
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i found the derivative. 4(t-1)^3 3(t+5)^2
is it just \[(t-1)^4(t+5)^2\] ?
with the chain rule those are not the derivatives...

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Other answers:

yes but instead of 2 as an exponent its 3. i know im suppose to = 0 then im not sure what to do.
wait that is right, t is single power, sorry.
okay so i did it wrong can we review step by step
just to be clear, it isnt the two equations plus one another, its them two multiplied by one another?
yes
kk
so i get.... 4(t-1)^3 3(t+5)^2
(t-1)^3 (t+5)^2 (17+7 t)
how?
I'm trying to figure that out, I used Wolframalpha to check my work and it got something different.
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 \[\huge (t-1)^3(t+5)^2(7t+17)=0\] \[\huge t=1, t=-5, t=-\frac{17}{7}\]
\[s(t)=y\] To find the values of y, sub in the values of t into the original equation.
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.
how did you get to this step: =(t−1)3(t+5)2[4t+20+3t−3]

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