anonymous
  • anonymous
Find the critical numbers of f(x) = x^(4/5)(7x-28)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[f(x)=x^{\frac{4}{5}}(7x-28)^2\] right?
anonymous
  • anonymous
Yep! :)
anonymous
  • anonymous
I know I have to find where the derivative = 0 or does not exist... Things just get a little bit messy when I derive it

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anonymous
  • anonymous
it is a huge mess, that is why
anonymous
  • anonymous
Lol XD
anonymous
  • anonymous
one 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
  • anonymous
Ohh right!
anonymous
  • anonymous
you 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
  • anonymous
it 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
  • anonymous
turns out the numerator is \[98(7x^2-36x+32)\] and you can set that equal to zeros and solve via the quadratic formula
anonymous
  • anonymous
Yep I used chain rule and product rule and I got this... \[f'(x) = \frac{ 4 }{ 5 }x ^{-1/5}(7x-28)^2+14x ^{4/5}(7x-28)\]
anonymous
  • anonymous
nope
anonymous
  • anonymous
Oh how did you get that?
anonymous
  • anonymous
oh wait how did you get the second part?
anonymous
  • anonymous
yeah i think that is wrong
anonymous
  • anonymous
Isn't it product rule?
anonymous
  • anonymous
oh no it isn't my mistake you are right
anonymous
  • anonymous
now here is what you have to do
anonymous
  • anonymous
in order to write it as one fraction, (since the first part is a fraction) multiply top and bottom by \(\sqrt[5]{x}\)
anonymous
  • anonymous
Oh so we multiply it on the second piece only?
anonymous
  • anonymous
rational 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
  • anonymous
\[\frac{4(7x-28)^2}{5\sqrt[5]{x}}+14\sqrt[5]{x^2}(7x-28)\] is what you are looking at
anonymous
  • anonymous
to 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
  • anonymous
the numerator will be \[4(7x-28)^2+14x(7x-28)\] which you can do the algebra with to get a quadratic to solve
anonymous
  • anonymous
Ahh 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
  • anonymous
damn i made a typo, but you can probably spot it
anonymous
  • anonymous
Hmmm... where?
jim_thompson5910
  • jim_thompson5910
This is what I'm getting \[\large f(x) = x^{4/5}(7x-28)^2\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28)\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28) {\color{red}{\times \frac{5x^{1/5}}{5x^{1/5}}}}\] \[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+\frac{70x(7x-28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{4(7x-28)^2+70x(7x-28)}{5x^{1/5}}\] \[\large f \ '(x) = \frac{98(x-4)(7x-8)}{5x^{1/5}}\]
anonymous
  • anonymous
Oh the 5 was missing!! Thanks so much @jim_thompson5910 ! :)
jim_thompson5910
  • jim_thompson5910
no problem
anonymous
  • anonymous
Expanding it, I get huge numbers, is that how it's supposed to be??
jim_thompson5910
  • jim_thompson5910
set it equal to 0 and solve for x
jim_thompson5910
  • jim_thompson5910
namely the numerator
anonymous
  • anonymous
How did you get the last line @jim_thompson5910 ?
anonymous
  • anonymous
lol i was missing the five too doh!
anonymous
  • anonymous
also os has be acting up, i kept trying to send the wolfram line http://www.wolframalpha.com/input/?i=x^%284%2F5%29%287x-28%29^2
jim_thompson5910
  • jim_thompson5910
4(7x-28)^2 + 70x(7x-28) 2(7x-28)*[2(7x-28) + 35x] ... factor out 2(7x-28) 2(7x-28)*(14x-56 + 35x) 2(7x-28)*(49x - 56) 2[7(x-4)]*[7(7x-8)] 98(x-4)(7x-8)
anonymous
  • anonymous
Ohhh ok! Yeah I was confused on how to simplify that! XD Geee lol
anonymous
  • anonymous
Then I have to simplify the 2 parentheses right?
jim_thompson5910
  • jim_thompson5910
yeah set it equal to 0 and use the zero product property
anonymous
  • anonymous
Ok im gonna give that a try! :D
anonymous
  • anonymous
Thanks so much @jim_thompson5910 and @satellite73 !!! :D
jim_thompson5910
  • jim_thompson5910
you're welcome

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