Locate the absolute extrema of the function (if any exist) over the indicated variables.

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Locate the absolute extrema of the function (if any exist) over the indicated variables.

Mathematics
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This is how I solved a...
I found the derivative of the function: x(4-x^2) I set it to zero... x=0 4-x^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?

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your derivative is just a little off
also 4-x^2=0 gives you x=2 or x=-2
"your derivative is just a little off" What should it be? "also 4-x^2=0 gives you x=2 or x=-2" Yeah, that makes sense. I forgot about that rule.
you should have used chain rule
\[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]
I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o
Now that I know that. How could I find the critical points for that derivative?
\[f'(x)=\frac{-x}{\sqrt{4-x^2}}\] is this what you have for f'?
Not even close unfortunately
you 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 4-x^2=0 so you are right your critical number are -2,0,2
wait what?
did you not understand how I got... \[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]
Yes, I actually solved it wrongly, and some how managed to get the right values. This is how I originally solved it...
\[\sqrt{4-x^2}\]\[(4-x^2)^\frac{1}{2}\]\[\frac{1}{2}(4-x^2)(2x)\]\[x(4-x^2)\] I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.
Then I got... x=0 4-x^2=0 or x=2 , x=-2
Ok yeah but are you understand why this is \[f'(x)=\frac{-x}{\sqrt{4-x^2}}\] ?
I know how you get to \[-x(4-x^2)^\frac{1}{2}\] is what you have just a simplified version of that?
well 1/2-1 is -1/2 not 1/2
Sorry, typo
\[-x(4-x^2)^\frac{-1}{2}=\frac{-x}{(4-x^2)^\frac{1}{2}} \text{ by law of exponents }\]
\[x^{-n}=\frac{1}{x^n}\]
There are so many laws I need to re-memorize.
But yes I do recognize what you did.
anyways 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
1?
\[f(x)=\sqrt{4-x^2} \\ f(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?\]
question d has 1 as an endpoint...
Oh okay, I was just looking at a
I just wanted to get all the evaluation done with
we can look at a first ... I want to get through evaluating the function at the critical numbers and any endpoints
anyways can you evaluate those 3 above...
x=2 would end up being 0 x=-1 would also be 0 x=0 would be 2 x=1 would be sqrt(3)
That was supposed to say -2 not -1
f(-2)=0 f(0)=2 f(2)=0 f(1)=sqrt(3) great work
so look at the outputs above
0,2,0
what is biggest number there
2
so that is your absolute max
0,2,0 and of these numbers 0 is the absolute min
That makes sense, but how would I represent that max as coordinates?
which this makes sense... because the graph is the top part of a circle... |dw:1444500172873:dw|
well you already found the coordinates
f(x)=y means (x,y) is a coordinate on f
for example we said f(2)=0 so that means (2,0) is a coordinate on f
Oh so max would be 0,2
(0,2) would be max ordered pairs have ( ) around them
|dw:1444500273762:dw|
so could (-2,0) and (2,0) both be min values?
yep
Because b has a parentheses instead of a bracket, would it be solved for differently?
now 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
Why can't we include 0?
[-2,0) means we have everything between -2 and 0 including -2 but not including 0
|dw: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
[ means included ( means to that point but the point isn't included itself
Correct?
see 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
Ah okay
and for C it would just be solving for 0?
brackets means we include the endpoint parenthesis means we don't include the endpoint
we 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]
So only the max would exist in that particular problem.
yep
Should i just state that it cannot be determined or is there a specific term I should use?
I think the word choice I used was good "does not exist" |dw:1444500908510:dw| this graph on [1,2) looks like: |dw:1444500919987:dw|
Thank you so much for your help! :D
what did you get for b again
max: does not exist min: (-2, 0)
great
and what about d
just a second
max: sqrt(3) min: does not exist That may or may not be right.
did you still want the coordinates or will the y value just be okay ?
like you can say the max is sqrt(3) but if your teacher prefers coordinates then you should say (1,sqrt(3))
Yeah that's what I'll put
ok ok that sounds great to me
excellent! It's surprising how simple this stuff really is compared to how difficult it seems at times
It 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
I do the worst on trig identities personally

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