Question

The graph of f(x) with a domain of [-3,3] is composed of line segments as seen in the graph below (attached)

a.find f'(x) and sketch the graph. (ill sketch the graph, just need help finding f'(x))
b. Name the x-coordinate for each point of discontinuity of f'(x) over (-3,3)

Answer 1

a. f'(x) is the slope of f(x). from ur sketch, there are 3 lines, do u know what the slope is for each line? They are the values for f'(x).

Answer 2

no I don't know

Answer 3

slope of a straight line between (x1, y1) and (x2, y2) = (y1 - y2) / (x1 - x2)

does that help?

Answer 4

@superdavesuper I got:
1st line- 1
2nd- -5/3
3rd- 5/2

Answer 5

hmmmm show me ur steps for the 1st line plz....the answer is not right.....

Answer 6

-3,1&-2, 2
1-2/-3+2=-1/-1=1

Answer 7

@superdavesuper

Answer 8

but from ur pic, the first 2 pts are (3,2) and (1, -3) right?

Answer 9

im going left to right? its the smallest line

Answer 10

Ahhhh okok sorry my mistake

So those are the value of f'(x) between [-3,3] ok?

Answer 11

yeah so f'(x) is 1-5/3+5/2?

Answer 12

Nope, f'(x)=1 for x=-3 to -2
f'(x)=? for x=-2 to 1
f'(x)=?? for x=1 to 3

Plz fill out ? and ?? up there

Answer 13

?=-5/3
??=5/2

Answer 14

Very good - that's about it for part a.

Now what u think of part b? look at the values above, do u notice anything funny about them?

Answer 15

wait so still on part a... I just graph each point and connect the lines?

Answer 16

3 separate horizontal lines for f'(x)

Answer 17

ok now onto part b

Answer 18

Read my last 2 posts n see if u notice something strange about f'

Answer 19

well I still don't completely understand how to graph them. I tried using (-1,-1), (-5,3), and (5,2) but theres only two lines

Answer 20

sorry 4 my bad drawing but thats how f'(x) should look

Answer 21

ok

Answer 22

I have to go now. thanks for your help!

Answer 23

draw f'(x) out later n u will see where it is discontinued :)