Question

1. Use Calc. to determine the critical points, local extrema, inflection points, and intervals where f(x) is concave up or down.

f(x) = x^4 + 2x^3 - 1

Answer 1

whoa... so u just want...everything... then...?

Answer 2

OK @amelapal ... so first of all, what are the turning points...?

Answer 3

well the extrema is the hardest part for me I don't know why :-(

Answer 4

okay so the critical points are when the function is equal to zero right?

Answer 5

**the function's derivative

Answer 6

sorry, hit send and saw ur post, my bad

Answer 7

its alright just caught my mistake!

Answer 8

so in this case the derivative would 4x^3 + 6x^2?

Answer 9

perfect

Answer 10

so one critical point can be zero right?

Answer 11

yep, but sub back in to initial eqn prove

Answer 12

*to prove

Answer 13

plug zero back into the original function? it gives you negative one, what do I do with that?

Answer 14

@Jack1

Answer 15

Sooo you've got zero, any others? :)

Answer 16

No! For some reason I can't find any other numbers!

Answer 17

\[\Large\bf\sf 0=4x^3+6x^2\]Factoring an x^2 out of each,\[\Large\bf\sf 0=x^2(4x+6)\]
x^2=0 gives us our critical point at x=0.
How bout the other factor?

Answer 18

4x + 6 = 0
4x = -6
x= -6/4 ?

Answer 19

Ok good, which simplifies to x=-3/2 or x=-1.5

Answer 20

Those are the critical points of the function.
So what's next?
Determining max and min?

Answer 21

that would be another critical point?

Answer 22

wait I get it! okay yes finding max and min now

Answer 23

you plug both 0 in the original function and get -1
and plug -6/4 and get -2.6875

Answer 24

No we don't want to do that! :Oo

Answer 25

So like ummm

Answer 26

:o

Answer 27

`critical points` / `stationary points` / `equilibrium points`
they go by many different names, depending on what field of work you're in.

It's a special place where the function is not `increasing` nor is it `decreasing`.

The function is at the bottom of a valley,
or at the top of a hill.

So what we want to do is,
check out what's happening on each side of the critical point.

Answer 28

Here is a nice little shortcut,
We only care about the `sign` of the result.
\[\Large\bf\sf f'(x)=x^2(4x+6)\]See how we have a square that we factored out?
That will ALWAYS be positive, no matter what we plug into the function.
So we can ignore that part.\[\Large\bf\sf f'(x)=(positive)(4x+6)\]Let's plug -2 into the derivative function.

Answer 29

\[\Large\bf\sf f'(-2)=(positive)(-8+6)\]\[\Large\bf\sf f'(-2)=(positive)(negative)\]\[\Large\bf\sf f'(-2)=negative\]

Answer 30

Negative tells us that the function is `decreasing` on the left side of x=-1.5

Answer 31

Does this process make sense .. a lil bit maybe? :o

Answer 32

We have 2 more areas we have to check.
We want to know what's happening on the right side of x=-1.5
and then on the right side of x=0.

Answer 33

brb i need some chocolate milk +_+

Answer 34

You look at the stuff, think :U
lemme know if it's too crazy.

Answer 35

it makes total sense! but brb I have to get ready for school now -_-!

Answer 36

oh heh :3

Answer 37

apologies @amelapal , laptop issues, @zepdrix finished it up though