Question

Can someone help me find the infl points and all relative extrema of this function, -> y=x^3/3+6x^2+2

Answer 1

Inflection points are where the second derivative is equal to zero

Answer 2

after i get those x coordinates, would i plug them back into the equation to get the full points?

Answer 3

Yes. Plugging the x value back into the equation will then give you the corresponding y value of the inflection point.

Answer 4

thank you very much (:

Answer 5

You're welcome. Remember, the extrema are the global and local maximums and minimums. I would suggest plotting the original curve to find those.

Answer 6

I got (-13, 282) (-5,109) and (6, 289) as the maxes

Answer 7

no mins

Answer 8

OK. How did you find the extrema?

Answer 9

I got the critical number of the first derivative and used intervals

Answer 10

then i tested numbers between those intervals and all three of them were greater than 0. so they were maxes

Answer 11

What do you mean by the 'critical number' ?

Answer 12

I found the x values when i set the first derivative equal to zero

Answer 13

and then i set up intervals from those x values to test them

Answer 14

from -inf to +inf

Answer 15

What are your solutions in x when you set the first derivative equal to zero?

Answer 16

I got 0 and -12

Answer 17

the intervals i chose were (-inf, -12) (12,0) and then (0, +inf)

Answer 18

OK, those are the correct intervals.

First, since you only have two roots to the first derivative, there are only two extrema.

Second, none of the three points you said the extrema were at "I got (-13, 282) (-5,109) and (6, 289) as the maxes" have 0 or -12 as an x coordinate.

You need to take the roots of the first derivative and plug them into the original equation to find the corresponding y coordinate

Answer 19

I got the inflection points at 0,1 and -6, 145

Answer 20

But those are the inflection points, how would I find the actual extrema? I know the intervals are correct but I don't understand how those wouldn't be the correct points for the rel maxes

Answer 21

So, there is only one inflection point in this problem. Typically, I like to find the extrema first and then find the inflection points. Sorry for steering you in a different direction.

So you have the roots to when the first derivative is equal to zero. Those tell you the location of your extrema. Plug those values back into the original equation and you will find the ordered pair for the location of each extrema.

Since you only have two roots to the first derivative, you only have two extrema.

Let me know what the coordinates are

Answer 22

No its fine. One second, figuring it out

Answer 23

Well, testing the intervals between those two points 0,1 and -6, 145 I have 0,1 as a max and -6,145 as a max as well

Answer 24

because the points between both are greater than zero when i plug them in.

Answer 25

Oh wait, one second, I am supposed to plug in -12 instead of -6 right? haha oops. one sec

Answer 26

OK. We aren't to testing the intervals yet. Just to make sure we are on the same page the original equation is \[y=1/3 x^3+6x^2+2\]

Answer 27

yes

Answer 28

Yes. you should be plugging in x=0 and x=-12

Answer 29

okay, one second

Answer 30

well, 0,1 is one of the extremas

Answer 31

I didn't get that. Check it again

Answer 32

and, -12, 289 is another

Answer 33

So now, 0,1 and -12, 289 are relative maxes

Answer 34

apparently their are no mins

Answer 35

I think the last term on the right hand side of your equation is a 1. Is it supposed to be a 2?

Answer 36

oh yes, its a 1. my mistake

Answer 37

Ahh, that's why we are getting different answers.

OK, those are your extrema. Now, lets find if they are maximums or minimums. Take the first derivative of the equation and plug in x values contained in the three intervals that you had. Do you get a positive derivative or a negative derivative for each interval?

Answer 38

Between the intervals i will test, -13 , -6 and 3 respectively

Answer 39

When I tested those, they were all positive

Answer 40

I got a negative derivative for when x=-6. It is not possible for there to be extrema and all derivatives to be positive unless you have a discontinuous function. This function is a polynomial so it is continuous on the interval (-infinity,infinity). That is a clue right away that there is something wrong.

Try an easier number like x=-1. The easier the better :)

Answer 41

I will test that, one sec

Answer 42

When i tested -1 I got a positive answer, 20/3 > 0

Answer 43

What is your first derivative?

Answer 44

The first derivative I got was x(x+12)

Answer 45

Ok. -1(-1+12)=-11 <0. This one is negative.

Answer 46

Oh, so you plug them into the derivative, not the actual equation?

Answer 47

In that case, If i were to plug -13, -1 and 3 respectively back into the first derivative, I would get -13,13 as a max -1,-11 as a min and 3,45 as a max as well

Answer 48

Yes. To find out what kind of extrema they are, you need to test the derivative to see if the slope is increasing or decreasing.

Answer 49

and then 0,1 would be an inflection point? their is only one you said. how is that?

Answer 50

No, (0, 1) is an extrema. Remember, y'[x]=0 is for the extrema and y''[x]=0 is for the inflection points.

Lets figure out whether the extrema are max or min before we get to the inflection point. So what do you think about the extrema?

Answer 51

Oh okay. I think the extrema is (-13,13) (-1,-11) (3,45) and (0,1) as you said

Answer 52

-1, -11 is a min

Answer 53

Remember, there are only two extrema that correspond to the roots of y'[x]=0.

Answer 54

Okay, so the those are the extremas thus far. 0, 1 is also a max

Answer 55

would -12, 289 be an extrema as well?

Answer 56

Yes, (0,1) and (-12, 289) are the extrema. Do you know how to tell if they are maximums or minimums?

Answer 57

Yes, by plugging in points in between, and they are both maxes

Answer 58

They can't both be maximums. Remember, the derivative of y[x] from (-infinity,-12)>0, y'[x] from (-12,0)<0, and y'[x] from (0,infinity)>0

Answer 59

So then -12, 289 must be a min.

Answer 60

No. So the derivative is increasing from -infinity to -12 and then starts decreasing at -12. So that means a rough slope of the curve is doing the same thing and it looks like an upside down parabola with the vertex at (-12, 289). This means it's a maximum.

What about (0, 1)?

Answer 61

0,1 would be a relative min.

Answer 62

after plugging in 0.5 in between

Answer 63

Yes. Very Good. I like to use 0, 1 and -1 if I can when plugging in to find extrema because they are the easiest to work with.

So do you still have a question about the inflection point or are you all set?

Answer 64

Well, I get the inflection points after finding the second derivative, setting it to zero and then plugging the x values to get the corresponding points. And those inflection points were 0,1 and -6,145. But you said I could only have one, so which would one would they be?

Answer 65

OK, what is your second derivative?

Answer 66

The second derivative is x(x+6) i believe

Answer 67

I have y''[x]=12+2x

Answer 68

or 2(6+x)

Answer 69

oh yes, sorry haha, it does yield the same answers for the x values, so I would 0,1 and -6, 145

Answer 70

since the greatest power of x (the highest order) in the second derivative is one, there is only one root. Solve for x
y''[x]=0=12+2x

Answer 71

x=-6

Answer 72

so -6, 145

Answer 73

Yes. That is your inflection point, or where the second derivative equals zero.

Answer 74

thank you (:

Answer 75

You're welcome. Keep up the good work.

Answer 76

Thanks again and have a nice night/ or day (: haha