We were given those question today on the test. And I am still abit confused by them.
1) I basically "proved" the thing works by using y=x^3, and p=0
And in that function. b=1, c=0, d=0, e=0. And I proved that the formula works.
But is that the right way to solve it?
2)I got stuck and we can't use logarithm.
3)Yeah, I got stuck. 3 is done. Wow I am a idoit.
Can any one show me the steps for 2 and 3?
EDIT: All done.
PM me for solution if you are interested (I don't want my teacher to google this lol).
JuvLupin
2009-11-03 23:00:07
Regarding question two, I'm not to sure if this helps but from my knowledge (started limits not too long ago) if substitution and cancelling don't work then you would approach "0" from the "+" and "-". However, both directions would be the same value since the "x" value would multiply by itself. So say we sub 0.25 (or -0.25) into the limit then we'd get 10^(-1/0.0625). Which would then become 10^(-16), basically 1/10^16. So as the "x" value get infinitely close to "x", the limit from both sides are zero, therefore the limit is zero.
yeah, thats what i though. But I am not too sure if we were allowed to sub in.
Basically all of this thing is done.
THANKS ALL (JuvLupin)
Nava2
2009-11-04 00:00:06
For 1, derive it. Then, rearrange the slope formula for p. Sub p in as y, I think it would work out…
so
err.
I tried using f(x+h)-f(x)
but since we have x^3+0x^2+0x+0. i ended up with p=3x^2 with no way of knowing b,c,d.
Nava2
2009-11-04 09:03:05
From: Main_Ftw
so
err.
I tried using f(x+h)-f(x)
but since we have x^3+0x^2+0x+0. i ended up with p=3x^2 with no way of knowing b,c,d.
Rearrange the second equation for p, then sub in y as p, and it should give you some of the variables out.