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

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

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

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

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

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

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

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

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

MoonlitFate
 one year ago
Best ResponseYou've already chosen the best response.0Also 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.

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

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

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

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

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

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

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

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

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

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

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