Question

Hello I am stuck with this one:
(f(x-h)-f(x)) all over h

Answer 1

It is just
\[
\frac{f(x-h)-f(x)}{h}
\]You can't do anything with this unless you have some more definitions...?

Answer 2

h does not equal zero

Answer 3

oh and f(x) is x^2-2x+1

Answer 4

oh fun derivative equation

Answer 5

If
f(1)=1^2-3*1+1
f(y)=y^2-3y+1
f(z+1)=(z+1)^2-3(z+1)+1

f(x-h)=?

Answer 6

yea that is what i got stuck on

Answer 7

follow the patterns i just showed you...

Answer 8

oops the 3 should be a 2

Answer 9

ooooh thanks

Answer 10

yes thats what the parenthesis means

Answer 11

I forgot how to do that over the summer

Answer 12

if you dont get 2x-2, try it again

Answer 13

ok thank you very much

Answer 14

wait how did you do that I am not even close

Answer 15

i cheated, i know the short cut for deriving equations

the actual thing should be a limit
\[\lim_{x \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]
\[\lim_{x \rightarrow 0}\frac{(x+h)^2-2(x+h)+1-(x^2-2x+1)}{h}\]
\[\lim_{x \rightarrow 0}\frac{x^2+2xh+h^2-2x-2h+1-x^2+2x-1}{h}\]
\[\lim_{x \rightarrow 0}\frac{2xh+h^2-2h}{h}\]
\[\lim_{x \rightarrow 0}2x+h-2\]

Answer 16

oh thanks but that other guy said it was just 2x-2

Answer 17

oh it should be the limit h->0 not x-> 0
oops

Answer 18

solve for the limit h->0

Answer 19

... I haven't done limits yet

Answer 20

ok then substitute h=0

Answer 21

2x+h−2
h=0
2x+0-2
2x-2

Answer 22

oh wow i forgot all that stuff

Answer 23

but then its over zero

Answer 24

also h is not equal to zero

Answer 25

it says