Calculus: Absolute Minima and Maxima

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[kickit246]

@BeckyGoesMoo If it is a polynomial then it always will be but I'm pretty sure if u have something like (x^2 - 7x)/(x + 3) You can see that the denominator will be zero if x = -3. Having the denominator = zero = undefined. Things of that nature. I think there are a few more cases like that but can't think of them at the moment. I'm actually terrible at calculus but figured I could help you with that question. And no problem, I love to help when I can.

[BeckyGoesMoo]

@kickit246 btw, how can you tell if f'(x) is defined everywhere?

[BeckyGoesMoo]

@kickit246 Thank you! That totally helped! :D

[kickit246]

@BeckyGoesMoo Sorry thats meant to be f(-1) = 3(-1)^4 - 8(-1)^3 + 6(-1)^2 + 3 the first time i write it out there.

[kickit246]

@BeckyGoesMoo Shes taking the critical numbers that she found as well as the end points of the interval given in the question and substituting them in for X. Example f(1) = 3(1)^4 - 8(1)^3 + 6(1)^2 + 3. You do the exponents first, which is -1^4 = 1, -1^3 = -1, etc. So the question then becomes 3(1) - 8(-1) + 6(1) + 3 = 20. The negative symbols in -8(-1) make it positive so it adds instead of subtracts. She just did that but with all the critical points and interval numbers given. Hope this helps

[BeckyGoesMoo]

How did she know to add all the numbers to find f(-1)?

and why is f(0) = 3? I have no idea how she found f(1) and f(2) as well.

Can somebody please explain to me how she found those numbers?

[Myersaws]

Awesome! My only question is how would you do this if your graph is linear?

Your f'(x) would be a number like 4, or 2/5. So, you're not going to be able to find any values by setting it equal to zero.

And a linear graph has no point where it is not defined. So, what would you do in that case?

[badboyzkikz]

thanks. that has helped a lot!

[witeyify]

The best cri/abs min/max vid. so far..thx B-)

[funh3ll]

my god this chick is hot.....jkjk

thanks for posting, this is very helpful/