Question

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

Answer 1

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

Answer 2

Yep! :)

Answer 3

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

Answer 4

it is a huge mess, that is why

Answer 5

Lol XD

Answer 6

one place it will not exist is at \(x=0\) since the derivative of \(x^{\frac{4}{5}}\) is
\[\frac{4}{5}x^{-\frac{1}{5}}=\frac{4}{5}\frac{1}{\sqrt[5]{x}}\]

Answer 7

Ohh right!

Answer 8

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

Answer 9

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

Answer 10

turns out the numerator is
\[98(7x^2-36x+32)\] and you can set that equal to zeros and solve via the quadratic formula

Answer 11

Yep I used chain rule and product rule and I got this... \[f'(x) = \frac{ 4 }{ 5 }x ^{-1/5}(7x-28)^2+14x ^{4/5}(7x-28)\]

Answer 12

nope

Answer 13

Oh how did you get that?

Answer 14

oh wait how did you get the second part?

Answer 15

yeah i think that is wrong

Answer 16

Isn't it product rule?

Answer 17

oh no it isn't my mistake
you are right

Answer 18

now here is what you have to do

Answer 19

in order to write it as one fraction, (since the first part is a fraction) multiply top and bottom by \(\sqrt[5]{x}\)

Answer 20

Oh so we multiply it on the second piece only?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

the numerator will be \[4(7x-28)^2+14x(7x-28)\] which you can do the algebra with to get a quadratic to solve

Answer 25

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?

Answer 26

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

Answer 27

Hmmm... where?

Answer 28

This is what I'm getting

\[\large f(x) = x^{4/5}(7x-28)^2\]

\[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28)\]

\[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+14x^{4/5}(7x-28) {\color{red}{\times \frac{5x^{1/5}}{5x^{1/5}}}}\]

\[\large f \ '(x) = \frac{4(7x-28)^2}{5x^{1/5}}+\frac{70x(7x-28)}{5x^{1/5}}\]

\[\large f \ '(x) = \frac{4(7x-28)^2+70x(7x-28)}{5x^{1/5}}\]

\[\large f \ '(x) = \frac{98(x-4)(7x-8)}{5x^{1/5}}\]

Answer 29

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

Answer 30

no problem

Answer 31

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

Answer 32

set it equal to 0 and solve for x

Answer 33

namely the numerator

Answer 34

How did you get the last line @jim_thompson5910 ?

Answer 35

lol i was missing the five too doh!

Answer 36

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

Answer 37

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)

Answer 38

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

Answer 39

Then I have to simplify the 2 parentheses right?

Answer 40

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

Answer 41

Ok im gonna give that a try! :D

Answer 42

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

Answer 43

you're welcome