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What answer did you get?

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

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

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

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?

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.

err +/- rather.

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

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