You are not logged in.
Question: Find the equation of the tangent to the curve at the given point. f(x) = 1/ square root of (x-1) at the point (2,1)
Using m = lim f(a+h) - f(a) / h how do we solve this? (Bold represents square root)
h->0
= lim f(2+h) - 1 / h
h->0
= 1/((2+h)-1) -1 / h
= 1/(1+h) -1 / h
Now to get rid of the 1/(1+h), we must get rid of the radical but also the numerators fraction. This is where I always get really confused and screw up .
Any help with this will be greatly appreciated.
we begin with
filling in for f(a) and f(a +h)
adding the two fractions in the numerator, and multiplying above and below by 1/h we obtain
now try the following, multiply the limit by the fraction shown on the right (the fraction is equal to 1)
from here, you should be able to work it out yourself.
note that
the reason we do this, is because the numerator had the form (a - b), where a and b were both the square root of something, thus if we multiply above and below by (a +b), the numerator becomes (a - b)*(a+b) = a^2 - b^2, the squaring gets rid of the radicals.
whenever you have limits involving radicals, this trick is usually needed to solve it.
Last edited by mikau (2008-02-06 16:58:52)
A logarithm is just a misspelled algorithm.
Offline
I am still having difficulty with getting rid of the fractions. The radical part I understand.
I go for help to my teacher and he tells me I can't do common denominators. Than he tells me to answer 1/2 + 1/3....which doesn't really help me and hes just trying to make me feel dumber and dumber. As you can see he's a great teacher and he makes math really fun
okay, we got up to here
doing the multiplications we obtain
note the h's above and below will cancel so you get
now the quotient rule for limits says the limit of a quotient (or fraction) is equal to the limit of the numerator divided by the limit of the denominator if and only if the denominator is nonzero. I think you will now find that the limit of the denominator is not zero. This means you can basically replace h with zero and evaluate it directly. So take the above limit, replace h with zero, and simplify!
Last edited by mikau (2008-02-06 17:03:26)
A logarithm is just a misspelled algorithm.
Offline
Wouldnt
be
Last edited by LuisRodg (2008-02-06 16:38:06)
Offline
argh! mess up! thanks, for the heads up, LuisRodg
(edit) FIXED!
Last edited by mikau (2008-02-06 16:55:47)
A logarithm is just a misspelled algorithm.
Offline
No problem.
Also, instead of using \sqrt( ) use \sqrt{ } so the root covers everything. Just makes it more readable. Just a suggestion.
Last edited by LuisRodg (2008-02-06 16:52:56)
Offline
No problem.
Also, instead of using \sqrt( ) use \sqrt{ } so the root covers everything. Just makes it more readable. Just a suggestion.
vs
oh THATS why thats happening! Thanks, that was really annoying.
A logarithm is just a misspelled algorithm.
Offline