Question

Simplify this :

Answer 1

\[\frac{ [1-(1+3h+3h^2+3h^3)]-(2) }{ h }\]

Answer 2

precalculus? or no?

Answer 3

yes it is

Answer 4

you should do 1 - (1 + 3h + 3h^2 + 3h^3) first

Answer 5

ohh okay

Answer 6

do i just get rid of the first round bracket?

Answer 7

@lgbasallote

Answer 8

correct. Distribute the - through [1 -1*( everything inside parenthesis) ]

Answer 9

oh so you mutiply everything by -1 ?:S

Answer 10

@cruffo

Answer 11

sorry had to step away..

Answer 12

it's alright :P

Answer 13

correct:
\[\frac{ [1-(1+3h+3h^2+3h^3)]-(2) }{ h } = \frac{ [1-1-3h-3h^2-3h^3]-2 }{ h }\]

Answer 14

but I'm curious: did you originally have to do 1-(1+h)^3 in the top?

Answer 15

well i have this equation: \[y=1-x^3\] and i need to find the average slope of the line passig through A (-1,2) and Δx=h

Answer 16

so that was what i came up with

Answer 17

I think before you go any further, there is something that need to be corrected.

Answer 18

ohh what ?

Answer 19

Here x = -1, so you want
\[f(-1+h) = 1-(-1+h)^3 = 1-[(-1)^3 + 3(-1)^2h + 3(-1)h^2 + h^3\]
\[ = 1-(-1+3h-3h^2+h^3)\]
So the top would be
\[\frac{ [1-(-1+3h-3h^2+h^3)]-(2) }{ h } = \frac{ [1+1-3h+3h^2-h^3]-2 }{ h }\]

Answer 20

then the -2 would cancel out with 2

Answer 21

and we would factor out the h ?

Answer 22

you got it.

Answer 23

so m= \[3+3h-h^2\]

Answer 24

yep.

Answer 25

oh wrong signs

Answer 26

hah. right. those pesky negatives...

Answer 27

but my text book says the final answer is: \[-h2+3h-3\]
:S :S

Answer 28

just rearrange the terms in decending order in h

Answer 29

ohh it's the same thing .. :$ LOL i thought it was different haha :$

Answer 30

thank-you so much :D !

Answer 31

np!

Answer 32

if part b says to calculate slope for h= 10,2,1,1/2 do i find the limits for his @cruffo ?

Answer 33

Sorry for the delay.

Answer 34

Should just be able to plug in the numbers for h in you previous answer, \[\large -h^2 + 3h - 3\]

Answer 35

thanks :D