anonymous
  • anonymous
Find critical points of: A(t)=t^5-5t^4+20t^3-17 I have an answer, but I think it's incorrect. Help, please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
What answer did you get?
anonymous
  • anonymous
First I found derivative: A'(t)=5t^4-20t^3+60t^2 Then factored out 5t^2 for: 5t^2(t^2-4t+12) So I got x=0, but used quadratic equation and got 4+-sqrt(-32) all over 2
anonymous
  • anonymous
I am concerned because I don't think there is supposed to be a negative under the sqrt sign.

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anonymous
  • anonymous
That just means that you won't have any real roots from that factor.
anonymous
  • anonymous
Okay. So If I'm finding absolute extrema, what do I do in this case? And my work looks correct to you?
anonymous
  • anonymous
So your only (real) critical points will be when t = 0.
anonymous
  • anonymous
Yes it is correct.
anonymous
  • anonymous
Okay.
anonymous
  • anonymous
Does that mean that the quadratic part is where it DNE?
anonymous
  • anonymous
DNE?
anonymous
  • anonymous
does not exist
anonymous
  • anonymous
The function exists for any real input. You just won't ever have a zero from that quadratic over the real numbers.
anonymous
  • anonymous
So look at what your derivative is doing about 0 and see whether you have a min, a max, or an inflection point.
anonymous
  • anonymous
Okay. Thank you! So is t=0 my only critical point?
anonymous
  • anonymous
For real inputs, yes.
anonymous
  • anonymous
So I'm on the closed interval [-4,3] I evaluate A(0), A(-4), and A(3), correct?
anonymous
  • anonymous
Sure, 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
  • anonymous
Hmm...How do I know that?
anonymous
  • anonymous
Well, if t is some number close to 0, but negative what would your derivative be?
anonymous
  • anonymous
Wait - do I evaluate the derivative or the original equation?
anonymous
  • anonymous
We're looking at the derivative to find if 0 is a local min max.
anonymous
  • anonymous
I thought I evaluate the function. Sorry I don't understand what I evaluate the derivative for.
anonymous
  • anonymous
And the derivative would be positive because t^2 is positive and (12 +/1 a very small number) is positive.
anonymous
  • anonymous
err +/- rather.
anonymous
  • anonymous
So my absolute max is 361 @ t=3 and absolute min is -3601 @ t=-4 right? (I'm only looking for max extrema.)
anonymous
  • anonymous
Ok, 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
  • anonymous
Since 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
  • anonymous
Oh. Okay! I get the y=x^3 part. My professor talked about that.
anonymous
  • anonymous
Have a nice day. I'm off to the store! =)

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