## mathcalculus Locate all critical points of s(t) = (2 t - 3)^(2) (8 - 2 t)^(3)? 7 months ago 7 months ago

Ok is this a calc problem?

2. mathcalculus

yes

3. mathcalculus

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

5. mathcalculus

critical points are x

6. mathcalculus

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

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

8. SithsAndGiggles

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

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

10. SithsAndGiggles

@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. mathcalculus

@SithsAndGiggles we used the product AND chain rule right?

Yep

13. SithsAndGiggles

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

14. SithsAndGiggles

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

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. mathcalculus

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