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

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