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anonymous
 3 years ago
Locate all critical points of s(t) = (2 t  3)^(2) (8  2 t)^(3)?
anonymous
 3 years ago
Locate all critical points of s(t) = (2 t  3)^(2) (8  2 t)^(3)?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok is this a calc problem?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0SO your critical points are max, min, and points of inflection, right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0critical points are x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we have to solve for x... to find the y's and that gives us our point

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right but when the question says critical points...In calculus that is usually the maximum, minimum an/or points of inflection.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[s(t)=(2t3)^2(82t)^3\\ \Rightarrow~s'(t)=4(2t3)(82t)^36(2t3)^2(82t)^2\\ \Rightarrow~s'(t)=(2t3)(82t)^2\left[4(82t)6(2t3)\right]\\ \Rightarrow~s'(t)=(2t3)(82t)^2(5020t)\] It should be fairly easy to find when \(s'(t)=0\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That would locate your max and minimums but not your points of inflection. @SithsAndGiggles

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad, usually when a problem asks for critical points, it's referring to the critical points of \(f(x)\). Points of inflection for \(f(x)\) are critical points of \(f'(x)\), not \(f(x)\), so I would think that's the assumption to be used here.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles we used the product AND chain rule right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Product, chain, power (not necessarily in that order).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad, but that doesn't mean critical points can't be inflection points. Depends on the function.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Correct .. I just wanted to know if she would need to go on and get the 2nd derivative to find the points of inflection as well or just the first derivative critical points.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@Paynesdad oh sorry, no i was only looking for the critical points.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Sounds like you have a good handle on this problem...Nice work @SithsAndGiggles and @mathcalculus
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