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anonymous
 5 years ago
Find critical points of: A(t)=t^55t^4+20t^317
I have an answer, but I think it's incorrect. Help, please!
anonymous
 5 years ago
Find critical points of: A(t)=t^55t^4+20t^317 I have an answer, but I think it's incorrect. Help, please!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What answer did you get?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First I found derivative: A'(t)=5t^420t^3+60t^2 Then factored out 5t^2 for: 5t^2(t^24t+12) So I got x=0, but used quadratic equation and got 4+sqrt(32) all over 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I am concerned because I don't think there is supposed to be a negative under the sqrt sign.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That just means that you won't have any real roots from that factor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. So If I'm finding absolute extrema, what do I do in this case? And my work looks correct to you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your only (real) critical points will be when t = 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does that mean that the quadratic part is where it DNE?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The function exists for any real input. You just won't ever have a zero from that quadratic over the real numbers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So look at what your derivative is doing about 0 and see whether you have a min, a max, or an inflection point.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. Thank you! So is t=0 my only critical point?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0For real inputs, yes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So I'm on the closed interval [4,3] I evaluate A(0), A(4), and A(3), correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure, but you don't really have to bother with 0 since it's pretty easy to see that it's an inflection point and not a local min/max.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hmm...How do I know that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, if t is some number close to 0, but negative what would your derivative be?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Wait  do I evaluate the derivative or the original equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We're looking at the derivative to find if 0 is a local min max.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought I evaluate the function. Sorry I don't understand what I evaluate the derivative for.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And the derivative would be positive because t^2 is positive and (12 +/1 a very small number) is positive.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So my absolute max is 361 @ t=3 and absolute min is 3601 @ t=4 right? (I'm only looking for max extrema.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, you know that if the derivative is positive the curve is sloping upward. If the derivative is negatve it slopes down. If you have a local max then the derivative will be 0 at that point, but just before that point it will be positive, and after it will be negative. If you have a local min then the derivative will be 0, but just before the point it will be negative, and just after that it will be positive. If you have an inflection point (think y=x^3 around 0) you will have the derivative is 0, but it doesn't change signs from one side of the point to the other. So when looking for extrema you want to take into consideration local mins/maxes and the values at the end points.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Since in our case the derivative is positive just before 0 and positive just after 0 we can see that it's just an inflection point and not a possible min/max.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. Okay! I get the y=x^3 part. My professor talked about that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Have a nice day. I'm off to the store! =)
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