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

Calculus1
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So the function h(x) = \[\sqrt{8-6x^2}\] , is that right? By critical points, do you mean relative maxima / minima?
yes
i mean x and y

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Other answers:

critical points..
Oh, okay. First of all, do you know how to take the derivative of h(x)?
so far im here: -12x/ 2 radical 8-6x^2
The derivative of h(x) is: \[\frac{ -6x }{ \sqrt{8-6x ^{2}} }\]
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.
So, you're going to set f'(x) to 0 and solve for x.
okay makes sense... then to find the critical points i equal that to zero... and just solve for x?
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?
\[\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \] \[(\sqrt{8-6x ^{2}}) *\frac{ -6x }{ \sqrt{8-6x ^{2}} } = 0 \sqrt{8-6x ^{2}}\] \[-6x = 0\] \[x=0\]
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.
I got zero as well... not how do i find y
i plugged in zero into the normal question... and got 2 radical 2
answer: (0,2 radical 2) is wrong://
Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(
its okay can we do after we find the x
the y..
Plug in the critical numbers into the f(x).
i plug in 0 in the original function? to find y right
Yes, sorry. Working it out over here, too. ^^;
\[2\sqrt{2}\]...
i keep getting wrong :(
?
\[f(0) = \sqrt{8-6(0)^2} = \sqrt{8}= 2\sqrt{2}\]
I was actually getting kinda confused, and I ended up graphing it to check.

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