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anonymous
 one year ago
Locate the absolute extrema of the function (if any exist) over the indicated variables.
anonymous
 one year ago
Locate the absolute extrema of the function (if any exist) over the indicated variables.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is how I solved a...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I found the derivative of the function: x(4x^2) I set it to zero... x=0 4x^2=0 or x=2 I then plugged the following back into the original function: 2, 0, 2 And I received: 0, 2, 0 So I concluded that my max = 2 and my min = 0. Assuming that what I did was correct, how could I figure out the exact coordinates of my min and max without looking at a graph?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1your derivative is just a little off

freckles
 one year ago
Best ResponseYou've already chosen the best response.1also 4x^2=0 gives you x=2 or x=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0"your derivative is just a little off" What should it be? "also 4x^2=0 gives you x=2 or x=2" Yeah, that makes sense. I forgot about that rule.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you should have used chain rule

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[((4x^2)^\frac{1}{2})'=\frac{1}{2}(4x^2)^{\frac{1}{2}} \cdot (4x^2)'\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now that I know that. How could I find the critical points for that derivative?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f'(x)=\frac{x}{\sqrt{4x^2}}\] is this what you have for f'?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not even close unfortunately

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you can find the critical numbers by finding when it is 0 or when it does not exist (these numbers should be in the domain of the original function though) so you still have to solve the equations x=0 and 4x^2=0 so you are right your critical number are 2,0,2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you not understand how I got... \[((4x^2)^\frac{1}{2})'=\frac{1}{2}(4x^2)^{\frac{1}{2}} \cdot (4x^2)'\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I actually solved it wrongly, and some how managed to get the right values. This is how I originally solved it...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{4x^2}\]\[(4x^2)^\frac{1}{2}\]\[\frac{1}{2}(4x^2)(2x)\]\[x(4x^2)\] I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I got... x=0 4x^2=0 or x=2 , x=2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Ok yeah but are you understand why this is \[f'(x)=\frac{x}{\sqrt{4x^2}}\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know how you get to \[x(4x^2)^\frac{1}{2}\] is what you have just a simplified version of that?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well 1/21 is 1/2 not 1/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[x(4x^2)^\frac{1}{2}=\frac{x}{(4x^2)^\frac{1}{2}} \text{ by law of exponents }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[x^{n}=\frac{1}{x^n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There are so many laws I need to rememorize.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But yes I do recognize what you did.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways you have the right critical numbers x=2,0,2 so we need to evaluate f(2),f(0),f(1),f(2) I put f(1) because we need to evaluate it later anyways

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[f(x)=\sqrt{4x^2} \\ f(2)=\sqrt{4(2)^2}=\sqrt{44}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1question d has 1 as an endpoint...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay, I was just looking at a

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I just wanted to get all the evaluation done with

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we can look at a first ... I want to get through evaluating the function at the critical numbers and any endpoints

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways can you evaluate those 3 above...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0x=2 would end up being 0 x=1 would also be 0 x=0 would be 2 x=1 would be sqrt(3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That was supposed to say 2 not 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1f(2)=0 f(0)=2 f(2)=0 f(1)=sqrt(3) great work

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so look at the outputs above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what is biggest number there

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so that is your absolute max

freckles
 one year ago
Best ResponseYou've already chosen the best response.10,2,0 and of these numbers 0 is the absolute min

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes sense, but how would I represent that max as coordinates?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1which this makes sense... because the graph is the top part of a circle... dw:1444500172873:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well you already found the coordinates

freckles
 one year ago
Best ResponseYou've already chosen the best response.1f(x)=y means (x,y) is a coordinate on f

freckles
 one year ago
Best ResponseYou've already chosen the best response.1for example we said f(2)=0 so that means (2,0) is a coordinate on f

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so max would be 0,2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1(0,2) would be max ordered pairs have ( ) around them

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444500273762:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so could (2,0) and (2,0) both be min values?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because b has a parentheses instead of a bracket, would it be solved for differently?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1now we can look at b we have [2,0) we only care about f(2) and f(0) here but we cannot actually include what happens at x=0 f(2)=0 and f(0)=2 is what we found earlier of 0 and 2 the smallest is 0 so your absolute min is at x=0 and we will not include an absolute max for this question because we cannot include what happens at x=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why can't we include 0?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1[2,0) means we have everything between 2 and 0 including 2 but not including 0

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444500569732:dw this function on [2,0) looks like: dw:1444500587078:dw my quarter of a circle was horribly drawn I blame it on the cat licking me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0[ means included ( means to that point but the point isn't included itself

freckles
 one year ago
Best ResponseYou've already chosen the best response.1see the absolute max doesn't actually exist there is a hole at x=0 because the domain didn't allow us to include what happens at x=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and for C it would just be solving for 0?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1brackets means we include the endpoint parenthesis means we don't include the endpoint

freckles
 one year ago
Best ResponseYou've already chosen the best response.1we already looked at a and said the absolute mins were (2,0) and (2,0) and the absolute max was (0,2) c would be the same answer except the absolute mins would not exist since we have (2,2) and not [2,2]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So only the max would exist in that particular problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Should i just state that it cannot be determined or is there a specific term I should use?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I think the word choice I used was good "does not exist" dw:1444500908510:dw this graph on [1,2) looks like: dw:1444500919987:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much for your help! :D

freckles
 one year ago
Best ResponseYou've already chosen the best response.1what did you get for b again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0max: does not exist min: (2, 0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0max: sqrt(3) min: does not exist That may or may not be right.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1did you still want the coordinates or will the y value just be okay ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like you can say the max is sqrt(3) but if your teacher prefers coordinates then you should say (1,sqrt(3))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah that's what I'll put

freckles
 one year ago
Best ResponseYou've already chosen the best response.1ok ok that sounds great to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0excellent! It's surprising how simple this stuff really is compared to how difficult it seems at times

freckles
 one year ago
Best ResponseYou've already chosen the best response.1It is really not that bad... You do have to train yourself back on your algebra though.. The calculus I think is easier than the algebra :p

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I do the worst on trig identities personally
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