Question

Find first and second derivative, the intervals of which the function is increasing,decreasing,concave up, and concave down, and locate any relative maxima,minima

Answer 1

\[f(x)=4x^3-6x^2-72x+20\]

Answer 2

@rational

Answer 3

please help me

Answer 4

Were you able to find the first derivative? :o

Answer 5

i think its 12x^2-12x-72

Answer 6

\[\Large\rm f'(x)=12x^2-12x-72\]Mmm ok good.
So to find increasing/decreasing we need critical points first.
Those are the spots where it can change from increasing to decreasing, etc..
To find critical points we set the first derivative equal to zero and solve for x, yes?

Answer 7

\[\Large\rm 0=12x^2-12x-72\]What x values do you get? :)

Answer 8

so i take out 12(x^2-12x-72)

Answer 9

Careful! Make sure you take a 12 out of each term, not just the first!!\[\Large\rm 0=12(x^2-x-6)\]

Answer 10

okay

Answer 11

whats next @zepdrix

Answer 12

So you need to solve for x to find the critical points.
Divide each side by 12,\[\Large\rm 0=x^2-x-6\]Looks like it will factor, yes?

Answer 13

yes

Answer 14

(x-6)(x+1)

Answer 15

Hmm I don't think those work out correctly.
-6 and 1 will give us -5 for our middle term.
Try again silly! :U

Answer 16

Oppsies (x-5)(x-1)

Answer 17

?? :(

Answer 18

You need a -6,
So factor of -6...
-6 times 1 didn't work.

Other factors of -6? :o
-5 and -1 are not factors.

Answer 19

what?.....

Answer 20

Don't remember how to factor..? :(

\[\Large\rm 0=x^2-x-6\]Factor of -6 that add to -1 are going to be,\[\Large\rm -3\quad and\quad 2\]

Answer 21

\[\Large\rm 0=(x-3)(x+2)\]

Answer 22

it can also be -5 and -1 right?

Answer 23

No.
-5 times -1 gives 5.
Those are NOT factors of 6.

Answer 24

oh yeahhhhh....omg i cant believe i forgot

Answer 25

@zepdrix the it'll be x=3 and x=2 right?

Answer 26

3 and -2 yes?

Answer 27

oh yes

Answer 28

Ok good.. these are stationary points ( critical points ).
It's locations where the function is neither increasing nor decreasing.
Like the top of a hill, or the bottom of a valley.

We want to know what's happening on either side of these stationary points.

Answer 29

okay

Answer 30

So what we'll do is, we'll plug `test points` into the first derivative,

if the result is negative,
the function is decreasing there.

if the result is positive,
the function is increasing in that location.

Answer 31

Here is our derivative function with everything factored,\[\Large\rm f'(x)=(x-3)(x+2)\]

Answer 32

Let's test a point on the left side of -2.
How about -3?\[\Large\rm f'(-3)=12(-3-3)(-3+2)\]Does this end up being positive or negative?

Answer 33

Bahhh I missed the 12 the first time I pasted the derivative function.. my bad.

Answer 34

im confused

Answer 35

If the first derivative at this point is negative,
then we have decreasing.

Answer 36

So we plug this value (or any value smaller than -2) into our derivative function

Answer 37

so we substitute x to 3 and 2 with the derivative function?

Answer 38

No.
x=3 and x=-2 will give us zero in the derivative function.
We know that because they are critical points.

We want to know what is happening `near` our critical points.
That's why I chose a point near -2.

Answer 39

okay so we plug in -3 with what function

Answer 40

whats the derivative function again