Question

please help!!!! picture included

Answer 1

Intervals of increasing decreasing?
Ok so again, derivative, ya?

Answer 2

well i used product rule @zepdrix

Answer 3

(2/3)x^(-1/3)(5-2x)+(-2)(x^(2/3))

Answer 4

Looks good.
There is some tricky algebra we can do from here,
to clean it up.

This might be a little confusing.

Answer 5

What I would choose to do next is,
factor 2/3x^(-1/3) out of each term.\[\large\rm f'(x)=\frac{2}{3}x^{-1/3}\left[\qquad\qquad\qquad\qquad\right]\]And we'll be left with the stuff.

Answer 6

with some stuff* in the brackets

Answer 7

\[\large\rm f'(x)=\frac{2}{3}x^{-1/3}\left[(5-2x)+\qquad\qquad\right]\]

Answer 8

\[\large\rm f'(x)=\frac{2}{3}x^{-1/3}\left[(5-2x)+-3x\right]\]What do you think about that?
Too confusing?

Answer 9

it is a bit confusing can you explain please?

Answer 10

I'm taking -2x^{2/3} and dividing 2/3x^{-1/3} out of it.
That's what we mean by "factor".\[\large\rm \frac{-2x^{2/3}}{\frac{2}{3}x^{-1/3}}\]This is what gives us our second term in the brackets.

Answer 11

and we are left with only -3x if i understand correctly

Answer 12

Yes.
You flip the fraction in the bottom, and change it to multiplication,
and then apply exponent rule to the x's.\[\large\rm -2\cdot\frac{3}{2}x^{2/3-(-1/3)}\]

Answer 13

ok i see how you factored out 2/3x^(-1/3)

Answer 14

what do we do next? try to substitute a couple things?

Answer 15

Let's look for critical/stationary points.
First derivative tells us about slope.
critical points exist where the slope of the function is zero.
\[\large\rm f'(x)=\frac{2}{3}x^{-1/3}\left[5-2x-3x\right]\]
\[\large\rm 0=\frac{2}{3}x^{-1/3}\left[5-2x-3x\right]\]And solve for x.

Answer 16

Apply your `Zero-Factor Property`: \(\large\rm ab=0\quad\implies\quad a=0,\quad b=0\)

Answer 17

i do not know what the zero factor property is

Answer 18

Well I just stated what It is :)
You need to be comfortable with using it.

So applying it gives us,
\[\large\rm 0=\frac{2}{3}x^{-1/3},\qquad\qquad\qquad 0=\left[5-2x-3x\right]\]

Answer 19

When you have things multiplying together to give you zero,
then you set each individual factor to zero

Answer 20

i got a critical number at x=1

Answer 21

Ok great,
and we have another,
coming from the 2/3x^{-1/3} factor.

Answer 22

x=0

Answer 23

Good good good.
Let's apply our First Derivative Test.

Answer 24

We would like to know what is happening "around" these special points.

Answer 25

ive decided to plug in 1/2 into the first derivative

Answer 26

That will tell us what is happening "between" the critical points,
ok seems good!

Answer 27

i got 2.099

Answer 28

which means it is positive

Answer 29

so the slope is increasing

Answer 30

Ok great.