Question

Locate all absolute extrema using calculus.

Answer 1

\[x ^{1/3}(6-x)^{2/3}\]

Answer 2

@Kainui :)

Answer 3

(oh, that is f(x)= [the statement])

Answer 4

Find your derivative? :U

Answer 5

Product Rule to the rescueee

Answer 6

it looks..so complex. sigh. okaaay i'll product rule xP

Answer 7

what's it?
fg'+gf' ? right?

Answer 8

ya -_-

Answer 9

just a moment! sorry!

Answer 10

:D

Answer 11

\[x^{1/3}\frac{ 2 }{ 3 }(6-x)^{-1/3}+(6-x)^{2/3}1/3x^{-2/3}\]

Answer 12

...probably could have simplified that

Answer 13

Woops, forgot chain rule on the first part, ya?

Answer 14

oh inside? inthe (6-x)? isn't that just 1?

Answer 15

The derivative of (x-6) is 1, yes.
But the derivative of (6-x) is -1.
:)

Answer 16

*Because the negative is in front of the x.

Answer 17

oou so that entire first part I just put a negative in front

Answer 18

\[\large\rm f(x)=x ^{1/3}(6-x)^{2/3}\]Yes, ok that looks better.\[\large\rm f'(x)=-\frac{2}{3}x^{1/3}(6-x)^{-1/3}+\frac{1}{3}x^{-2/3}(6-x)^{2/3}\]

Answer 19

These next couple of steps are going to be a pain, but oh well.
We're going to get rid of the negative powers by putting stuff in the denominator,
and then we'll need to get a common denominator and combine it all into a single fraction.

Answer 20

ai ai ai. Let's do it.

Answer 21

Ok ok do it on paper >.<
I'll type it out cause it's gonna be a mess lol.

See if you can come up with the same thing.

Answer 22

Step one, getting rid of negative exponents:
\[\large\rm f'(x)=-\frac{2x^{1/3}}{3(6-x)^{1/3}}+\frac{(6-x)^{2/3}}{3x^{2/3}}\]

Answer 23

Step two, common denominator:\[\large\rm f'(x)=-\frac{2x^{1/3}\color{royalblue}{x^{2/3}}}{3\color{royalblue}{x^{2/3}}(6-x)^{1/3}}+\frac{(6-x)^{2/3}\color{royalblue}{(6-x)^{1/3}}}{3x^{2/3}\color{royalblue}{(6-x)^{1/3}}}\]

Answer 24

Simplify further,\[\large\rm f'(x)=-\frac{2x}{3x^{2/3}(6-x)^{1/3}}+\frac{(6-x)}{3x^{2/3}(6-x)^{1/3}}\]and further,\[\large\rm f'(x)=\frac{-2x+(6-x)}{3x^{2/3}(6-x)^{1/3}}\]

Answer 25

I know I went through that kinda quick :O
Takeeee yer time.

Answer 26

I am like a salamander competing against Olympic swimmers. But I will arrive at the end! just a moment!

Answer 27

can I just simplify the top to -3x+6 at the end there?

Answer 28

Yes.

Answer 29

Look for critical points.
Find any? :)

Answer 30

yaay. k

Answer 31

uuuh 2?

Answer 32

2 is good.
we also have some weird critical points coming from the denominator.

Answer 33

The value we got from the numerator corresponds to where the function has `horizontal tangency`.

The values we get from the denominator will correspond to where the function has `vertical tangency`.

A lot of times, when you're doing this type of problem,
the values you get from the denominator will not even be in the domain of the function,
so you don't have to worry about it usually.

But that is not the case for this problem,
we DO need those points.

Answer 34

\[\large\rm 0=\frac{3(2-x)}{3x^{2/3}(6-x)^{1/3}}\]

\[\large\rm 0=3(2-x),\qquad\qquad\qquad 0=3x^{2/3}(6-x)^{1/3}\]So what other x values do we get?

Answer 35

0 xP and...6..? o.o

Answer 36

Yes, good.

Answer 37

So let's make a number line for our derivative, and then choose test points,