Question

Local max and min question!

Answer 1

Ahhh she's back!
Just couldn't stay away from the ole O'Study could ya?

Answer 2

haha i'm surprised you are brave enough to tackle another one of my questions!

Answer 3

That's a terrible nickname, O'study..? Pretend I didn't say that...

Answer 4

xD
although it is not allowing me to attach a file. sigh.

Answer 5

Hmm site been really laggy today :[

Answer 6

Gah.
well..it's 4x^3+3x^2-6x+8
local max?
local min?

Answer 7

I got that the critical points are -1 and 1/4 but that is not working for the max and min?

Answer 8

A polynomial??
Come on Mare.. you can do this one -_-

Answer 9

hmm

Answer 10

I think you get 1/2 for the positive critical point, ya?
Maybe check your work again.

Answer 11

\[\large\rm 4x^3+3x^2-6x+8\qquad D\to\qquad 12x^2+6x-6\]Looking for critical points,\[\large\rm 0=2x^2+x-1\]You can try to factor it, I think Quadratic Formula is easier from here though.

Answer 12

oh yes, I meant 1/2 not 1/4

Answer 13

Oh :o

Answer 14

Critical points: -1, 1/2

Answer 15

What'd you get for your extrema?
-1 is the max, ya?

Answer 16

yes but it doesn't like that answer D:

Answer 17

oou because it's asking for locals maybe??

Answer 18

Locals? Yes maybe.
Did they give you a closed interval?

Answer 19

Nope

Answer 20

Oh hehe.

Answer 21

We found the `location of the local minimum`, ya? ;)
Do you see what we didn't do?

Answer 22

uhm uhm. no haha what? D:

Answer 23

We found the `local minimum LOCATION, x=1/2`,
they want us to go a step further and give the `local minimum VALUE`.

Answer 24

ou. so we want to put those into the original function?

Answer 25

Yes. That will give us VALUES for our min and max.

Answer 26

so for
-1 = 13
1/2 = 25/4 ?

Answer 27

Yes, good job.

Answer 28

Excellent, thank you! :)

Answer 29

np