Question

Determine the intervals where the graph is concave up and concave down. y = x^3 - 3x^2 + 4x - 1

Answer 1

@wio can u help?

Answer 2

Find inflection points when \(f''(x) = 0\). When \(f''(x)>0\) it is concave up.

Answer 3

how do i do that?

Answer 4

Do you know how to get the second derivative?

Answer 5

yes

Answer 6

Find it.

Answer 7

okay

Answer 8

i get 6x - 6

Answer 9

Set it equal to 0 and solve.

Answer 10

ok

Answer 11

That will be an inflection point.

Answer 12

I got 1

Answer 13

so 1 is the inflection point

Answer 14

Write 6x-6>0 and solve. It is concave up when the second derivative is greater than 0
And yes. 1 is the inflection point.

Answer 15

so it will be concave

Answer 16

Concave up when second derivative >0
Concave down when second derivative <0

Answer 17

okay

Answer 18

Here is the graph:
http://www.wolframalpha.com/input/?i=y+%3Dx^3+-+3x^2+%2B+4x+-+1

Answer 19

Okay so the second derivative is > 0 so it will be concave up right?

Answer 20

in general, critical values are determined by =0, or undefined

Answer 21

ok

Answer 22

so its wrong?

Answer 23

you need to define an some intervals, you found that at x=0, the concavity changes

Answer 24

okay how do i do that?

Answer 25

what is the concavity for x < 0, what is it for x > 0 ?

Answer 26

well for x < 0 it concave down and for x > 0 it concave up

Answer 27

correct, then your interval of cave up is? (0,inf) right? or do we have another point of inflection looming out there?

Answer 28

well how about 1?

Answer 29

6x - 6 = 0, i got 1 after i solved

Answer 30

fine, use 1 then :) i misread the top part and thought yall deduced this to 0 :)

cave up: (1,inf)

what would you determine as an interval for cave down?

Answer 31

umm would it have negative on it? or

Answer 32

it would, since it goes to -inf.

so, cave down: (-inf,1)

Answer 33

okay