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help :( locate all critical points ( both types ) of
h(x)= radical 86*x^2
The critical point(s) is (are) .
 one year ago
 one year ago
help :( locate all critical points ( both types ) of h(x)= radical 86*x^2 The critical point(s) is (are) .
 one year ago
 one year ago

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MoonlitFateBest ResponseYou've already chosen the best response.0
So the function h(x) = \[\sqrt{86x^2}\] , is that right? By critical points, do you mean relative maxima / minima?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
critical points..
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
Oh, okay. First of all, do you know how to take the derivative of h(x)?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
so far im here: 12x/ 2 radical 86x^2
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
The derivative of h(x) is: \[\frac{ 6x }{ \sqrt{86x ^{2}} }\]
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
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.
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
So, you're going to set f'(x) to 0 and solve for x.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
okay makes sense... then to find the critical points i equal that to zero... and just solve for x?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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?
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
\[\frac{ 6x }{ \sqrt{86x ^{2}} } = 0 \] \[(\sqrt{86x ^{2}}) *\frac{ 6x }{ \sqrt{86x ^{2}} } = 0 \sqrt{86x ^{2}}\] \[6x = 0\] \[x=0\]
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.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.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
I got zero as well... not how do i find y
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i plugged in zero into the normal question... and got 2 radical 2
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
answer: (0,2 radical 2) is wrong://
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
its okay can we do after we find the x
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
Plug in the critical numbers into the f(x).
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i plug in 0 in the original function? to find y right
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
Yes, sorry. Working it out over here, too. ^^;
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i keep getting wrong :(
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
\[f(0) = \sqrt{86(0)^2} = \sqrt{8}= 2\sqrt{2}\]
 one year ago

MoonlitFateBest ResponseYou've already chosen the best response.0
I was actually getting kinda confused, and I ended up graphing it to check.
 one year ago
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