anonymous
  • anonymous
find the critical points: s(t)= (t-1)^4 (t+5)^3
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
i found the derivative. 4(t-1)^3 3(t+5)^2
anonymous
  • anonymous
is it just \[(t-1)^4(t+5)^2\] ?
anonymous
  • anonymous
with the chain rule those are not the derivatives...

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

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