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

1. anonymous

Ok is this a calc problem?

2. anonymous

yes

3. anonymous

4. anonymous

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

5. anonymous

critical points are x

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

@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.

11. anonymous

@SithsAndGiggles we used the product AND chain rule right?

12. anonymous

Yep

13. anonymous

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

14. anonymous

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

15. anonymous

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.

16. anonymous

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

17. anonymous

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

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