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!

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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
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What answer did you get?
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
I am concerned because I don't think there is supposed to be a negative under the sqrt sign.

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Other answers:

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

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