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

- anonymous

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

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- anonymous

##### 1 Attachment

- anonymous

This is how I solved a...

- 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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## More answers

- freckles

your derivative is just a little off

- freckles

also 4-x^2=0 gives you x=2 or x=-2

- 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

you should have used chain rule

- freckles

\[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]

- anonymous

I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o

- anonymous

Now that I know that. How could I find the critical points for that derivative?

- freckles

\[f'(x)=\frac{-x}{\sqrt{4-x^2}}\]
is this what you have for f'?

- anonymous

Not even close unfortunately

- 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

wait what?

- 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

Yes, I actually solved it wrongly, and some how managed to get the right values.
This is how I originally solved it...

- 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

Then I got...
x=0
4-x^2=0 or x=2 , x=-2

- freckles

Ok yeah but are you understand why
this is \[f'(x)=\frac{-x}{\sqrt{4-x^2}}\]
?

- 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

well 1/2-1 is -1/2 not 1/2

- anonymous

Sorry, typo

- freckles

\[-x(4-x^2)^\frac{-1}{2}=\frac{-x}{(4-x^2)^\frac{1}{2}} \text{ by law of exponents }\]

- freckles

\[x^{-n}=\frac{1}{x^n}\]

- anonymous

There are so many laws I need to re-memorize.

- anonymous

But yes I do recognize what you did.

- 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

1?

- freckles

\[f(x)=\sqrt{4-x^2} \\ f(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?\]

- freckles

question d has 1 as an endpoint...

- anonymous

Oh okay, I was just looking at a

- freckles

I just wanted to get all the evaluation done with

- freckles

we can look at a first
... I want to get through evaluating the function at the critical numbers and any endpoints

- freckles

anyways can you evaluate those 3 above...

- 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

That was supposed to say -2 not -1

- freckles

f(-2)=0
f(0)=2
f(2)=0
f(1)=sqrt(3)
great work

- freckles

so look at the outputs above

- freckles

0,2,0

- freckles

what is biggest number there

- anonymous

2

- freckles

so that is your absolute max

- freckles

0,2,0
and of these numbers 0 is the absolute min

- anonymous

That makes sense, but how would I represent that max as coordinates?

- freckles

which this makes sense...
because the graph is the top part of a circle...
|dw:1444500172873:dw|

- freckles

well you already found the coordinates

- freckles

f(x)=y means (x,y) is a coordinate on f

- freckles

for example we said f(2)=0 so that means (2,0) is a coordinate on f

- anonymous

Oh so max would be 0,2

- freckles

(0,2) would be max
ordered pairs have ( ) around them

- freckles

|dw:1444500273762:dw|

- anonymous

so could (-2,0) and (2,0) both be min values?

- freckles

yep

- anonymous

Because b has a parentheses instead of a bracket, would it be solved for differently?

- 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

Why can't we include 0?

- freckles

[-2,0)
means we have everything between -2 and 0
including -2 but not including 0

- 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

[ means included
( means to that point but the point isn't included itself

- anonymous

Correct?

- 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

Ah okay

- anonymous

and for C it would just be solving for 0?

- freckles

brackets means we include the endpoint
parenthesis means we don't include the endpoint

- 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

So only the max would exist in that particular problem.

- freckles

yep

- anonymous

Should i just state that it cannot be determined or is there a specific term I should use?

- 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

Thank you so much for your help! :D

- freckles

what did you get for b again

- anonymous

max: does not exist
min: (-2, 0)

- freckles

great

- freckles

and what about d

- anonymous

just a second

- anonymous

max: sqrt(3)
min: does not exist
That may or may not be right.

- freckles

did you still want the coordinates
or will the y value just be okay ?

- freckles

like you can say the max is sqrt(3)
but if your teacher prefers coordinates then you should say (1,sqrt(3))

- anonymous

Yeah that's what I'll put

- freckles

ok ok
that sounds great to me

- anonymous

excellent!
It's surprising how simple this stuff really is compared to how difficult it seems at times

- 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

I do the worst on trig identities personally

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