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find all the critical point of f(x) = 3x^(5/3) - 15x^(2/3)

Mathematics
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Critical points are points where the derivative is zero.
Of course, differentiate first :)
ok.. i got x=0, x=2

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Other answers:

let's see.... \[\Large \frac{d}{dx}\left(3x^{\frac53}-15x^{\frac23}\right)=5x^{\frac23}-10x^{-\frac13}\]
is my answer correct?
Try 0. the derivative doesn't exist, because you have... \[\Large 5x^{\frac23}-10x^{-\frac13}=5x^{\frac23}-\frac1{10x^{\frac13}}\] an x in the denmoinator.
so that will be disregarded right? so i will have 2 for the value of x
so yes, that is a critical point (sorry, I forgot to mention that points where the derivative does not exist are also critical points)
owh.. i see
so what next? how will i graph?
Hang on, I'm not sure about the 2 yet. Let's try equating the derivative to zero. \[\Large0=5x^{\frac23}-\frac1{10x^{\frac13}}\]\[\Large5x^{\frac23}=\frac1{10x^{\frac13}}\]\[\Large5x^{\frac23}\cdot 10x^{\frac13}=1\]\[\Large50x=1\] Are you sure it's x = 2? :P
can you factor out 5x^(1/3) from 5x^(2/3)−10x^(−1/3)?
You can... but why would you do that?
what will be the answer if you would factor out that one? 5x^(-1/3) i got a little confused
I think it leads nowhere :)
owh.. lol.. so x = 1/50?
yeah :)
what is the answer if i would take out 5x^(-1/3) from 5x^(2/3) i really am confused on what will be the outcome
Hang on, I may have made a mistake.
Okay... apparently, 2 was right.
sorry.
yehey :D so.. what to do next ?
just like that? You're not going to ask me where my error was? :/ You didn't spot it yet... means you may well commit the same... come on, a little challenge :P
\[\Large0=5x^{\frac23}-\frac1{10x^{\frac13}}\] I should not have put 10 in the denominator. Big mistake. Correct would be \[\Large0=5x^{\frac23}-\frac{10}{x^{\frac13}}\color{green}\checkmark \]
you shouldn't include 10 in the denominator.. it should be on the numerator :D
yeah that :) sorry about that. I might be getting drowsy.
And as far as I know... we're done.
that's ok.. :))
We already have the critical points.
then how to graph?
whoops... you'll need to consult one of the big guys for that :3 I'm rather terrible at graphing, you see...
owh.. that'll be okay then.. thanks for the help! :D
No problem :)

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