Question

HELPPPPPPP!!!! locate all critical points (both types) of
h(x)= radical 8-6*x^2

The critical point(s):

Answer 1

I FOUND THE X WHICH IS 0. THEN I TRIED TO FIND Y BY PLUGGING IN ZERO BUT IT IS INCORRECT... :(

Answer 2

i know the steps to this problem. which is find the derivative and i did...

Answer 3

then = to zero. and i did that too. and I goy x but then im not sure how i got the answer wrong....

Answer 4

It's hard to read the function when it's not formatted.
Does this look accurate?\[\large h(x)=\sqrt{8-6x^2}\]Everything under the square root like that?

Answer 5

yes

Answer 6

Hmm so we get something like this, yes?
\[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\]

So you determined that \(\large x=0\) is a critical point.
There is also another.

A critical point of the function also exists where \(\large f'(x)\) is undefined. (Assuming that value is included in the domain of \(\large f\).

Answer 7

okay.

Answer 8

yeah I determined one critical point, but what about y?

Answer 9

I'm saying that there is another critical point,\[\large \sqrt{8-6x^2}\]When this thing equals zero, we're dividing by zero, so f'(x) is undefined, DNE.
\[\large 0=\sqrt{8-6x^2}\]
Solve for x to find your other critical point(s).

Answer 10

As for your y values, if you want to, you can plug x=0 into your original function, to find the corresponding y value. That will allow you to write it as an ordered pair if you want.

Answer 11

yes i found 2 radical 2

Answer 12

and i wrote the answer like this... (0,2 radical 2)

Answer 13

Hmm I think you end up with \(\large x=\pm \dfrac{2}{\sqrt3}\)

Answer 14

(0,[2\sqrt{2}\])

Answer 15

how?

Answer 16

so x is not 0?

Answer 17

\[\large 0=8-6x^2\]\[\large 8=6x^2\]\[\large \frac{8}{6}=x^2 \qquad \rightarrow \qquad \frac{4}{3}=x^2\]\[\large x=\pm \sqrt{\frac{4}{3}}\qquad \rightarrow \qquad x=\pm \frac{2}{\sqrt3}\]

Answer 18

to find the critical points.. dont we find the derivative first?

Answer 19

Yes, this was our derivative,\[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\]We set the numerator equal to zero to find one critical point.
We need to also set the denominator equal to zero to find the other(s).
Which is what I was showing.

Answer 20

okay... then

Answer 21

we solve the denominator = to 0 but what about the radical we ignore it?

Answer 22

have you used webwork?

Answer 23

im not sure how to plug in the answer

Answer 24

(0,

Answer 25

The square root? I squared both sides. 0^2 = 0

Answer 26

no the answer; how do we write it if there a positive and a negative 2 radical 3.

Answer 27

I've used webassign, not webworks. Hmm

Is there a panel on the left somewhere where you can use tools like sqrt?

Answer 28

the critical points are:

Answer 29

ive tried every way but it's "Wrong"

Answer 30

yes sqrt( )

Answer 31

Write it as three separate critical points:
(0, sqrt(8))
(2/sqrt(3),0)
(-2/sqrt(3),0)

Does that work maybe? D:

Answer 32

Operands of '*' are not of compatible types

Answer 33

because its between of (0, sqrt(8)) (2/sqrt(3),0) (-2/sqrt(3),0)

Answer 34

hold on im so confused.... lol how did you get (0,8)

Answer 35

(0, sqrt8)?

Answer 36

We found that x=0 is a critical point.
If we plug x=0 back into the original function it gives us, \(\large f(0)=\sqrt8\) right? We can write that as an ordered pair (0,sqrt(8))

Answer 37

"Operands of '*' are not of compatible types"

I don't understand what you're saying :o

Answer 38

not me, thats what webworks says..

Answer 39

Does it tell you to list them as ordered pairs?
Usually you just list critical points as x=

Answer 40

Locate all critical points ( both types )
The critical point(s) is (are) .
(The answers are to be points. Use parentheses in your answer(s). If there are no critical points, enter none .)

Answer 41

now i tried to put the answer like this: (0,sqrt(8))U(2/sqrt(3),0)U(-2/sqrt(3),0) but it says: Left endpoint must be less than right endpoint

Answer 42

Hmm I'm not sure how your assignment wants them formatted.
I would guess that you just separate them with commas.

(0,sqrt(8)), (2/sqrt(3),0), (-2/sqrt(3),0)

Answer 43

correctttt :)

Answer 44

okay, so after we find x, how do we get the + and - 2 radical 3? i mean we plug in 0 to the original but i got 2 radical 2

Answer 45

i understand the 0 and sqrt 8 order pair...

Answer 46

why couldnt i write 0, 2 radical 2 as a critical point?

Answer 47

Scroll up to see my steps. I don't want to write them out again.

Set the denominator equal to zero, solve for x.

Answer 48

nvm but now... how about the +&- 2 radical 3.

Answer 49

We have an \(\large x^2\). When we take the root of a variable, we get two solutions, the positive and negative root.

Answer 50

i understand how you made the cal. but why did you ignore the radical and wrote it as 8-6x^2

Answer 51

yes ,right. we have a postive and negative. i understand that but why did you ignore the radical and wrote it as 8-6x^2? when the original one had a radical .

Answer 52

I didn't ignore the radical.\[\large 0=\sqrt{8-6x^2}\]Step 1, square both sides.

Answer 53

thank you so much! :)

Answer 54

thank you thank you! ^_^

Answer 55

i'll keep that the radicals need a + and - in mind.