Question

Locate all critical points of s(t) = (2 t - 3)^(2) (8 - 2 t)^(3)?

Answer 1

Ok is this a calc problem?

Answer 2

yes

Answer 3

@Paynesdad

Answer 4

SO your critical points are max, min, and points of inflection, right?

Answer 5

critical points are x

Answer 6

we have to solve for x... to find the y's and that gives us our point

Answer 7

Right but when the question says critical points...In calculus that is usually the maximum, minimum an/or points of inflection.

Answer 8

\[s(t)=(2t-3)^2(8-2t)^3\\
\Rightarrow~s'(t)=4(2t-3)(8-2t)^3-6(2t-3)^2(8-2t)^2\\
\Rightarrow~s'(t)=(2t-3)(8-2t)^2\left[4(8-2t)-6(2t-3)\right]\\
\Rightarrow~s'(t)=(2t-3)(8-2t)^2(50-20t)\]
It should be fairly easy to find when \(s'(t)=0\).

Answer 9

That would locate your max and minimums but not your points of inflection. @SithsAndGiggles

Answer 10

@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.

Answer 11

@SithsAndGiggles we used the product AND chain rule right?

Answer 12

Yep

Answer 13

Product, chain, power (not necessarily in that order).

Answer 14

@Paynesdad, but that doesn't mean critical points can't be inflection points. Depends on the function.

Answer 15

Correct .. 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.

Answer 16

@Paynesdad oh sorry, no i was only looking for the critical points.

Answer 17

Sounds like you have a good handle on this problem...Nice work @SithsAndGiggles and @mathcalculus