Question

Consider the graph g', the derivative of a function g.
http://i.imgur.com/uSYHrNQ.png (Note: Sorry if it looks slightly bad I had to draw it to the best of my ability.)
edit: http://i.imgur.com/b7cTF1O.jpg (This version shows the curves better)

Determine the following and justify briefly.
a) The critical numbers of g.
b) The interval(s) where g is increasing / decreasing.
c) The interval(s) where g is concave up / down.
d) The x co-ord of the points of inflection of g
e) The x co-ord of the relative extrema of g. (State if it is a maximum or minimum)
f) Roughly sketch g

Answer 1

it look like exactly same. you just make it bigger, right?

Answer 2

for a, b you can directly get the answer from the graph of g'
for c, you have to graph the graph of g''
for d,e, you have to graph the original function g getting from g'
it's quite hard to me. just give you idea/

Answer 3

The problem is i'm struggling with graphing g and g'' :\ so yeah. but i'm pretty sure:
a) = (-2,0) and (1,0)
b) (-inf,-2) decreasing
(-2,1) increasing
(1,inf) increasing

is that correct?

Answer 4

Are you sure about that? Because I thought how it worked was to get g from g' you look at the graph g' find the points where y=0 (so -2 and 1) and look if the points before after and in between those numbers are above or below the y axis. if they are below the graph is decreasing if they are above the graph is increasing.

Answer 5

ok, my bad, I take back my answer, I have to confirm the knowledge before giving out the right answer.

Answer 6

I need time to read my note. I 'll be back right after I figure out the answer, but don't worry . You have Mertsji here, He 's good enough for this question;

Answer 7

If the first derivative is 0 or undefined, there is a critical number.

Answer 8

If the graph of the first derivative is above the x axis (positive), the function is increasing. If it is below the x axis, the function is decreasing

Answer 9

@Mertsj no need to look up the information from the note, my note here, you :)

Answer 10

@Mertsj alright does that mean that this is correct?
a) = (-2,0) and (1,0)
b) (-inf,-2) decreasing
(-2,1) increasing
(1,inf) increasing

Answer 11

yes

Answer 12

Function is concave up if the first derivative is increasing
concave down, if it is decreasing.

Answer 13

@Mertsj so does that mean that (-inf,-1) concave up (-1,1) concave down (1,inf) concave up?

Answer 14

yes. Sorry for the delay...we're dealing with flooding here again today.

Answer 15

@Mertsj thats sad to hear. hopefully everything is alright wherever you are. What about d and e? What would I need to do there?

Answer 16

To find the inflection points, you should consider each critical point. Test a point on each side of the critical point. If the first derivative is changing from negative to positive at the critical point, that is a minimum. It is changes from positive to negative, it is a maximum. If neither, it is an inflection point.

Answer 17

How am I support to test the points if I do not have a function?

Answer 18

Look at the graph. This will be a visual test.

Answer 19

@Mertsj I'm not exactly sure but heres what I think:

x | f(x)
-3 | -inf
-1 | 2
0 | 1
2 | 2

so:
Negative - decreasing
Positive - increasing
Positive - increasing
Positive - increasing

Now I don't know what to do..

Answer 20

Are you talking about making the sketch of g now?

Answer 21

No i'm talking about:
d) The x co-ord of the points of inflection of g
e) The x co-ord of the relative extrema of g. (State if it is a maximum or minimum)

I don't understand it because I need co-ords.

Answer 22

The first derivative changes from negative to positive at -2. Therefore -2 is the x coordinate of a minimum point.
The first derivative is 0 at x = 1 but the first derivative is positive to the left of 1 and positive to the right of 1. Therefore 1 is not a maximum or a minimum. Therefore 1 is an inflection point. Therefore the x coordinate of the inflection point is 1

Answer 23

@Mertsj So wait does this mean that (-2,f(-2)) is my minimum, (1,f(1)) is my inflection point and there is no maximum?

Answer 24

Yes.

Answer 25

@Mertsj Alright I think I will be able to graph this on my own. Thank you so much for your help!

Answer 26

I hope so because I am terrible at sketching these things. And you are welcome.