Question

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

Answer 1

This is how I solved a...

Answer 2

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?

Answer 3

your derivative is just a little off

Answer 4

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

Answer 5

"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.

Answer 6

you should have used chain rule

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Not even close unfortunately

Answer 12

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

Answer 13

wait what?

Answer 14

did you not understand how I got...
\[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]

Answer 15

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

This is how I originally solved it...

Answer 16

\[\sqrt{4-x^2}\]\[(4-x^2)^\frac{1}{2}\]\[\frac{1}{2}(4-x^2)(2x)\]\[x(4-x^2)\]
I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.

Answer 17

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

Answer 18

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

Answer 19

I know how you get to \[-x(4-x^2)^\frac{1}{2}\] is what you have just a simplified version of that?

Answer 20

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

Answer 21

Sorry, typo

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

But yes I do recognize what you did.

Answer 26

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

Answer 27

1?

Answer 28

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

Answer 29

question d has 1 as an endpoint...

Answer 30

Oh okay, I was just looking at a

Answer 31

I just wanted to get all the evaluation done with

Answer 32

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

Answer 33

anyways can you evaluate those 3 above...

Answer 34

x=2 would end up being 0
x=-1 would also be 0
x=0 would be 2
x=1 would be sqrt(3)

Answer 35

That was supposed to say -2 not -1

Answer 36

f(-2)=0
f(0)=2
f(2)=0

f(1)=sqrt(3)

great work

Answer 37

so look at the outputs above

Answer 38

0,2,0

Answer 39

what is biggest number there

Answer 40

2

Answer 41

so that is your absolute max

Answer 42

0,2,0

and of these numbers 0 is the absolute min

Answer 43

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

Answer 44

which this makes sense...
because the graph is the top part of a circle...