Question

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!

Answer 1

What answer did you get?

Answer 2

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

Answer 3

I am concerned because I don't think there is supposed to be a negative under the sqrt sign.

Answer 4

That just means that you won't have any real roots from that factor.

Answer 5

Okay. So If I'm finding absolute extrema, what do I do in this case?
And my work looks correct to you?

Answer 6

So your only (real) critical points will be when t = 0.

Answer 7

Yes it is correct.

Answer 8

Okay.

Answer 9

Does that mean that the quadratic part is where it DNE?

Answer 10

DNE?

Answer 11

does not exist

Answer 12

The function exists for any real input. You just won't ever have a zero from that quadratic over the real numbers.

Answer 13

So look at what your derivative is doing about 0 and see whether you have a min, a max, or an inflection point.

Answer 14

Okay. Thank you! So is t=0 my only critical point?

Answer 15

For real inputs, yes.

Answer 16

So I'm on the closed interval [-4,3]
I evaluate A(0), A(-4), and A(3), correct?

Answer 17

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.

Answer 18

Hmm...How do I know that?

Answer 19

Well, if t is some number close to 0, but negative what would your derivative be?

Answer 20

Wait - do I evaluate the derivative or the original equation?

Answer 21

We're looking at the derivative to find if 0 is a local min max.

Answer 22

I thought I evaluate the function. Sorry I don't understand what I evaluate the derivative for.

Answer 23

And the derivative would be positive because t^2 is positive and (12 +/1 a very small number) is positive.

Answer 24

err +/- rather.

Answer 25

So my absolute max is 361 @ t=3
and absolute min is -3601 @ t=-4
right? (I'm only looking for max extrema.)

Answer 26

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.

Answer 27

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.

Answer 28

Oh. Okay! I get the y=x^3 part. My professor talked about that.

Answer 29

Have a nice day. I'm off to the store! =)