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

moonlitfate
 3 years 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)?

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

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

moonlitfate
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0So, you're going to set f'(x) to 0 and solve for x.

anonymous
 3 years 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?

anonymous
 3 years 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
 3 years 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
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got zero as well... not how do i find y

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i plugged in zero into the normal question... and got 2 radical 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0answer: (0,2 radical 2) is wrong://

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its okay can we do after we find the x

moonlitfate
 3 years ago
Best ResponseYou've already chosen the best response.0Plug in the critical numbers into the f(x).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i plug in 0 in the original function? to find y right

moonlitfate
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, sorry. Working it out over here, too. ^^;

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i keep getting wrong :(

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

moonlitfate
 3 years 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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