Find the critical numbers of f(x) = x^(4/5)(7x-28)^2

- anonymous

Find the critical numbers of f(x) = x^(4/5)(7x-28)^2

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- anonymous

\[f(x)=x^{\frac{4}{5}}(7x-28)^2\] right?

- anonymous

Yep! :)

- 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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## More answers

- anonymous

it is a huge mess, that is why

- anonymous

Lol XD

- 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

Ohh right!

- 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

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

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

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

nope

- anonymous

Oh how did you get that?

- anonymous

oh wait how did you get the second part?

- anonymous

yeah i think that is wrong

- anonymous

Isn't it product rule?

- anonymous

oh no it isn't my mistake
you are right

- anonymous

now here is what you have to do

- 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

Oh so we multiply it on the second piece only?

- 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

\[\frac{4(7x-28)^2}{5\sqrt[5]{x}}+14\sqrt[5]{x^2}(7x-28)\] is what you are looking at

- 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

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

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

damn i made a typo, but you can probably spot it

- anonymous

Hmmm... where?

- 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

Oh the 5 was missing!! Thanks so much @jim_thompson5910 ! :)

- jim_thompson5910

no problem

- anonymous

Expanding it, I get huge numbers, is that how it's supposed to be??

- jim_thompson5910

set it equal to 0 and solve for x

- jim_thompson5910

namely the numerator

- anonymous

How did you get the last line @jim_thompson5910 ?

- anonymous

lol i was missing the five too doh!

- 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

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

Ohhh ok! Yeah I was confused on how to simplify that! XD Geee lol

- anonymous

Then I have to simplify the 2 parentheses right?

- jim_thompson5910

yeah set it equal to 0 and use the zero product property

- anonymous

Ok im gonna give that a try! :D

- anonymous

Thanks so much @jim_thompson5910 and @satellite73 !!! :D

- jim_thompson5910

you're welcome

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