Question

help :( locate all critical points ( both types ) of
h(x)= radical 8-6*x^2
The critical point(s) is (are) .

Answer 1

So the function h(x) = \[\sqrt{8-6x^2}\] , is that right?

By critical points, do you mean relative maxima / minima?

Answer 2

yes

Answer 3

i mean x and y

Answer 4

critical points..

Answer 5

Oh, okay. First of all, do you know how to take the derivative of h(x)?

Answer 6

so far im here: -12x/ 2 radical 8-6x^2

Answer 7

The derivative of h(x) is: \[\frac{ -6x }{ \sqrt{8-6x ^{2}} }\]

Answer 8

The next step, is finding where f'(x) = 0 or is undefined, since that's going to find the critical points of f. And you can use those to form intervals to test for minima and maxima.

Answer 9

So, you're going to set f'(x) to 0 and solve for x.

Answer 10

okay makes sense... then to find the critical points i equal that to zero... and just solve for x?

Answer 11

oh so wait, i don't equal it to zero.... i plug in zero into x?? and from the original problem i plug int the x when i find that out?

Answer 12

\[\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \]
\[(\sqrt{8-6x ^{2}}) *\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \sqrt{8-6x ^{2}}\]
\[-6x = 0\]
\[x=0\]

Answer 13

Also f is undefined when:\[x = \pm \frac{ -2\sqrt{3} }{ 3 }\] and since there can't be a negative square root for the denominator, the domain of f is \[[x = \pm \frac{ -2\sqrt{3} }{ 3 }]\] , so you will use that along with 0 to make test intervals.

Answer 14

I got zero as well... not how do i find y

Answer 15

i plugged in zero into the normal question... and got 2 radical 2

Answer 16

answer: (0,2 radical 2) is wrong://

Answer 17

Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(

Answer 18

its okay can we do after we find the x

Answer 19

the y..

Answer 20

Plug in the critical numbers into the f(x).

Answer 21

i plug in 0 in the original function? to find y right

Answer 22

Yes, sorry. Working it out over here, too. ^^;

Answer 23

\[2\sqrt{2}\]...

Answer 24

i keep getting wrong :(

Answer 25

?

Answer 26

\[f(0) = \sqrt{8-6(0)^2} = \sqrt{8}= 2\sqrt{2}\]

Answer 27

I was actually getting kinda confused, and I ended up graphing it to check.