anonymous
  • anonymous
help :( locate all critical points ( both types ) of h(x)= radical 8-6*x^2 The critical point(s) is (are) .
Calculus1
katieb
  • katieb
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MoonlitFate
  • MoonlitFate
So the function h(x) = \[\sqrt{8-6x^2}\] , is that right? By critical points, do you mean relative maxima / minima?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i mean x and y

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

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