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anonymous
 one year ago
Find the critical numbers of f(x) = x^(4/5)(7x28)^2
anonymous
 one year ago
Find the critical numbers of f(x) = x^(4/5)(7x28)^2

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f(x)=x^{\frac{4}{5}}(7x28)^2\] right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know I have to find where the derivative = 0 or does not exist... Things just get a little bit messy when I derive it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is a huge mess, that is why

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0one place it will not exist is at \(x=0\) since the derivative of \(x^{\frac{4}{5}}\) is \[\frac{4}{5}x^{\frac{1}{5}}=\frac{4}{5}\frac{1}{\sqrt[5]{x}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you gotta use the chain rule and product rule to find the derivative or else expand the whole thing out and use the power rule repeatedly you still have to add up the fractions in any case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is a raft of boring algebra to get the derivative as one fraction i could walk you through it probably, but best idea is to cheat

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0turns out the numerator is \[98(7x^236x+32)\] and you can set that equal to zeros and solve via the quadratic formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep I used chain rule and product rule and I got this... \[f'(x) = \frac{ 4 }{ 5 }x ^{1/5}(7x28)^2+14x ^{4/5}(7x28)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh how did you get that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait how did you get the second part?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah i think that is wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't it product rule?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh no it isn't my mistake you are right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now here is what you have to do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in order to write it as one fraction, (since the first part is a fraction) multiply top and bottom by \(\sqrt[5]{x}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh so we multiply it on the second piece only?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0rational exponents are your friend when it comes to taking derivatives, but not when it comes to finding critical points or actually evaluating the derivative you need to get it out of exponential notation and write it as radicals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{4(7x28)^2}{5\sqrt[5]{x}}+14\sqrt[5]{x^2}(7x28)\] is what you are looking at

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0to put it over the same denominator multiply the second part top and bottom by \(\sqrt[5]{x}\) which is nice because the radical will go away

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the numerator will be \[4(7x28)^2+14x(7x28)\] which you can do the algebra with to get a quadratic to solve

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh I see!! So all I have to do now, is expand the numerator and set it = 0 and those are gonna be my critical numbers right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0damn i made a typo, but you can probably spot it

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1This is what I'm getting \[\large f(x) = x^{4/5}(7x28)^2\] \[\large f \ '(x) = \frac{4(7x28)^2}{5x^{1/5}}+14x^{4/5}(7x28)\] \[\large f \ '(x) = \frac{4(7x28)^2}{5x^{1/5}}+14x^{4/5}(7x28) {\color{red}{\times \frac{5x^{1/5}}{5x^{1/5}}}}\] \[\large f \ '(x) = \frac{4(7x28)^2}{5x^{1/5}}+\frac{70x(7x28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{4(7x28)^2+70x(7x28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{98(x4)(7x8)}{5x^{1/5}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh the 5 was missing!! Thanks so much @jim_thompson5910 ! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Expanding it, I get huge numbers, is that how it's supposed to be??

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1set it equal to 0 and solve for x

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1namely the numerator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you get the last line @jim_thompson5910 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lol i was missing the five too doh!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also os has be acting up, i kept trying to send the wolfram line http://www.wolframalpha.com/input/?i=x^%284%2F5%29%287x28%29^2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.14(7x28)^2 + 70x(7x28) 2(7x28)*[2(7x28) + 35x] ... factor out 2(7x28) 2(7x28)*(14x56 + 35x) 2(7x28)*(49x  56) 2[7(x4)]*[7(7x8)] 98(x4)(7x8)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh ok! Yeah I was confused on how to simplify that! XD Geee lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then I have to simplify the 2 parentheses right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah set it equal to 0 and use the zero product property

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok im gonna give that a try! :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much @jim_thompson5910 and @satellite73 !!! :D

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you're welcome
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