Question

Sketch the graph by finding where f is increasing and decreasing, where f is concave up and down, local extreme points, inflection points, and intercepts.

Answer 1

Of what function?

Answer 2

oh i'm sorry: f(x)=x^2/3(5/2-x)

Answer 3

to make it clearer, it's f(x)=[x^(2/3)][((5/2)-x)]

Answer 4

\[f(x)=x^{\frac{2}{3}}\left(\frac{5}{2}-x\right)\]

Answer 5

wow that looks professional :)

Answer 6

Ok, now the first thing you gotta do, is find the points that cross the x axis.
I used the button equation under the wrinting box.

Answer 7

so would you set the equation to 0?

Answer 8

Yes

Answer 9

so x=0, 5/2?

Answer 10

yes, now you need to see when this function is increasing and decreasing. Do you remember how to do that?

Answer 11

find crit pts where function is 0 and undefined, then make a number line?

Answer 12

Yes, but not where the function is 0, but where its derivative is 0.

Answer 13

oh ok. but this function can't be undefined, right?

Answer 14

Yes, but it is well defined in all reals.

Answer 15

i tried to get f'(x). is it 5/3(x)^(1/3) - 5(x)^(2/3)/3 ?

Answer 16

\[\frac{df}{dx}=\frac{2}{3}x^{-\frac{1}{3}}\left(\frac{5}{2}-x\right)-x^{\frac{2}{3}}=\frac{5}{3}x^{-\frac{1}{3}}-\frac{5}{3}x^{\frac{2}{3}}\]

Answer 17

cool. now i'm not sure how i would solve for x here...

Answer 18

Wait, did you understood what you missed when taking the derivative?

Answer 19

oh i thought it was the same, no?

Answer 20

Almost, you missed a minus in the first term, probably it was just a typo

Answer 21

no, i just put it in the denominator

Answer 22

So since the derivative tells if the function is increasing or decreasing, what you need now, is to get the intervals for wich the derivative is positive, negative, or zero.

Answer 23

Try to use the equation writer, its easier to understand

Answer 24

yeah this is actually where i got stuck, cuz i couldn't figure out what to do to to find the x-values.

Answer 25

Ok, now, to begin with, what happens at x=0?

Answer 26

it would equal 0?

Answer 27

no, there is a sqrt(x) in the denominator, so it is undefined. So now, the only way of the derivative give zero, would be that each term cancels the other so:\[\frac{5}{3}x^{-\frac{1}{3}}=\frac{5}{3}x^{\frac{2}{3}}\]

Answer 28

Not square root sorry, cubic root

Answer 29

would we use cross-multiplication from here?

Answer 30

thats one way, yes

Answer 31

ok let me try to solve this...

Answer 32

x=1, since 15x=15?

Answer 33

Yes, thats it. Now you need to analyze the behavior of the derivative around those critical points. See if it is positive or negative.
x<0 | x=0 | 0<x<1 | x=1 | x>1
+-? | ND | +-?? | y'=0 | +-??

Answer 34

Do you know how to do that?

Answer 35

i think so. would those values be plugged into f'(x)?

Answer 36

Yes, thats the function we are analyzing, but in those exact points we alredy know how it is, we want to know the ones with the ??

Answer 37

i forgot, how did we get the 0?

Answer 38

the 0 is where the first term os the derivative is not defined (ND)

Answer 39

Because of the division by 0

Answer 40

Do you want me to guide you through this step?

Answer 41

yes please, i would appreciate that :)

Answer 42

Ok, for the derivative to be positive for x>0, the positive term of the equation must be greater than the negative, so:\[\frac{5}{3}x^{-\frac{1}{3}}>\frac{5}{3}x^\frac{2}{3}\]\[1>x\]

Answer 43

plugging it in took me a while, but is it basically -, +, - for the number line?

Answer 44

Yes, thats it.

Answer 45

but i don't understand why you had to do that stuff up there. is that necessary to solve the problem?

Answer 46

Oooh, sorry, I didn't explain that

Answer 47

the derivative gives you the slope of the function, if it is negative at an interval, then in that interval f is decreasing, if it is positive f is increasing. So far our graph is like that just by looking at the derivatives and roots:The concavity is still unclear, but we will get there

Answer 48

Do you understand now?

Answer 49

Hey, I'll have to go for a few minutes, but I'll be back, try to find out what would be the next step to find the concavity

Answer 50

thats so cool! I see where you're going now. Ok, but please come back :)

Answer 51

Try to apply the same thinking on the concavity, think about f' increasing and decreasin (thats what concavity is). I will be back

Answer 52

Hey, I'm not going anymore.

Answer 53

oh what happened?

Answer 54

my sister got a ride...

Answer 55

oh thats cool.

Answer 56

So, what have you though about the next step?

Answer 57

just to make sure, for that previous part, would the local min be at x=0, and local max be at x=1?

Answer 58

Yes, thats correct

Answer 59

The only thing that is missing is the concavity and the brhavior of the function at infinity

Answer 60

Now, for f"(x), would the equation be \[-5/9(x) ^{-4/3}-10/9(x)^{-1/3}\]

Answer 61

Correct

Answer 62

solving for that, i got x =-1/2

Answer 63

Correct, and again undefined on x=0.

Answer 64

Do you know what to do next?

Answer 65

are they all concave down from \[(-\infty, -1/2), (-1/2, 0), (0, \infty)\]

Answer 66

oh, wait thats wrong.

Answer 67

What does it mean when the second derivative is zero?

Answer 68

Also, I always forget, is this concave down:Or this?