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terenzreignzBest ResponseYou've already chosen the best response.1
Critical points are points where the derivative is zero.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Of course, differentiate first :)
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
let's see.... \[\Large \frac{d}{dx}\left(3x^{\frac53}15x^{\frac23}\right)=5x^{\frac23}10x^{\frac13}\]
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
is my answer correct?
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
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.
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
so that will be disregarded right? so i will have 2 for the value of x
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
so yes, that is a critical point (sorry, I forgot to mention that points where the derivative does not exist are also critical points)
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
so what next? how will i graph?
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
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
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
can you factor out 5x^(1/3) from 5x^(2/3)−10x^(−1/3)?
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
You can... but why would you do that?
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
what will be the answer if you would factor out that one? 5x^(1/3) i got a little confused
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
I think it leads nowhere :)
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
owh.. lol.. so x = 1/50?
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
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
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Hang on, I may have made a mistake.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
Okay... apparently, 2 was right.
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
yehey :D so.. what to do next ?
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
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
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
\[\Large0=5x^{\frac23}\frac1{10x^{\frac13}}\] I should not have put 10 in the denominator. Big mistake. Correct would be \[\Large0=5x^{\frac23}\frac{10}{x^{\frac13}}\color{green}\checkmark \]
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
you shouldn't include 10 in the denominator.. it should be on the numerator :D
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
yeah that :) sorry about that. I might be getting drowsy.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
And as far as I know... we're done.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
We already have the critical points.
 10 months ago

terenzreignzBest ResponseYou've already chosen the best response.1
whoops... you'll need to consult one of the big guys for that :3 I'm rather terrible at graphing, you see...
 10 months ago

erosheaBest ResponseYou've already chosen the best response.1
owh.. that'll be okay then.. thanks for the help! :D
 10 months ago
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