## anonymous 4 years ago Simplify this :

1. anonymous

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

2. lgbasallote

precalculus? or no?

3. anonymous

yes it is

4. lgbasallote

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

5. anonymous

ohh okay

6. anonymous

do i just get rid of the first round bracket?

7. anonymous

@lgbasallote

8. cruffo

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

9. anonymous

oh so you mutiply everything by -1 ?:S

10. anonymous

@cruffo

11. cruffo

12. anonymous

it's alright :P

13. cruffo

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

14. cruffo

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

15. anonymous

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

16. anonymous

so that was what i came up with

17. cruffo

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

18. anonymous

ohh what ?

19. cruffo

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

20. anonymous

then the -2 would cancel out with 2

21. anonymous

and we would factor out the h ?

22. cruffo

you got it.

23. anonymous

so m= $3+3h-h^2$

24. cruffo

yep.

25. anonymous

oh wrong signs

26. cruffo

hah. right. those pesky negatives...

27. anonymous

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

28. cruffo

just rearrange the terms in decending order in h

29. anonymous

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

30. anonymous

thank-you so much :D !

31. cruffo

np!

32. anonymous

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

33. cruffo

Sorry for the delay.

34. cruffo

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

35. anonymous

thanks :D