# need help in completing a table

• Mar 12th 2008, 09:53 PM
sean1058
need help in completing a table
ok i already tried this but i'm not really sure if i did it right at all so i'm looking for some help. heres the problem.(spaces didnt work, i had to use underscores)

Use the function f(x)=2/x for the following.
a. Complete the table for x=2

h_|x+h| f(x+h) | f(x) | f(x+h)-f(x)/h
1_|___|_______|____|___________
.5_|___|______|____|___________
.1_|___|______|____|___________
.01|___|______|____|___________

now i completed the table by just plugging in 2 for x on all of the rows, along with the corresponding h value. so mine looks like this.

_h_|x+h|f(x+h)__|f(x)| f(x+h)-f(x)/h
_1_|_3_|__2/3__|_1_|-1/5
.5_|2.5_|__4/5__|_1_|-2/5
.1_|2.1_|20/21__|_1_|-1/2
.01|2.01|200/201|_1_|-1/2

is that the correct way to do it or am i missing something?
also there are a couple corresponding questions

b. Find f'(2) by a rule or formula
i got -1/4 for this

c. Is there a relationship between parts A and B? If so, what?
this is where i figured i did the table wrong, because i can't figure out what the relationship is. this is all for an important grade i could really use any help you can give me. thanks.
• Mar 12th 2008, 09:58 PM
TheEmptySet
$f(x)=\frac{2}{x}=2x^{-1}$

taking the derivative

$f'(x)=-2x^{-2}=\frac{-2}{x^2}$

then evaluate at 2

$f'(2)=\frac{-2}{2^2}=-\frac{1}{2}$
• Mar 12th 2008, 11:08 PM
sean1058
yeah i screwed up in typing that one sry.

im more concerned about the table. ive never had to do anything like that before, it just feels like i did it incorrectly.
• Mar 13th 2008, 05:59 AM
colby2152
Quote:

Originally Posted by sean1058
yeah i screwed up in typing that one sry.

im more concerned about the table. ive never had to do anything like that before, it just feels like i did it incorrectly.

You did the table correctly. You calculated the value of the derivative via formula wrong. Check out the latest in my Calculus Notes regarding derivatives. I show the tedious method of how to calculate, or estimate, them numerically.