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This is how I solved a...

your derivative is just a little off

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

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

wait what?

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.

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

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

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?

Why can't we include 0?

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

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

Correct?

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

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?

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 ?

Yeah that's what I'll put

ok ok
that sounds great to me

I do the worst on trig identities personally