Question

if f(x) = 2/x

then what is
[f(x+h)-f(x)] / h

Answer 1

Oh the reciprocal thing?
Ooo this one is a little stinker to simplify :) lol

Answer 2

i will attempt

Answer 3

k

Answer 4

so for f(x+h) does that mean 2/x+h
?

Answer 5

and for -f(x) : would that be -2/x

Answer 6

2/(x+h), yes
the brackets are kind of important, otherwise it looks like this \(\Large\rm \frac{2}{x}+h\)

Answer 7

mhm :d

Answer 8

lol
ok you're right I forgot about that.

Answer 9

ok so my whole equation will be {[2/(x+h)] - (2/x)} / h

Answer 10

@zepdrix

Answer 11

\[\Large\rm f(x)=\frac{2}{x}\]\[\Large\rm f(x+h)=\frac{2}{x+h}\]\[\Large\rm \frac{\frac{2}{x+h}-\frac{2}{x}}{h}\]
Mmm yes, looks good.

Answer 12

ok I remember seeing something like this before.

Answer 13

would I times the equation by h/1?

Answer 14

that way canceling out the bottom h?

Answer 15

So we have a couple of fractions in our large numerator.
How do we combine fractions?
We can't do the subtraction right now, since they don't have the same denominator.

Answer 16

h/1? Naw we can't just multiply by h/1 willy nilly.
Gotta multiply by h/h to keep the equation balanced.
But we don't actually want to do that here :d

Answer 17

oh.

Answer 18

oh ok. duh h/1 does not equal 1 anyways lol. (wish it were that easy) ok, so.... I think that I can start by taking care of the top two equation

Answer 19

Yes, do that.
They should simplify down into something nice.

Answer 20

so I have [ 2/(x+h)] and [2/x]
I believe their common denominator is x2+xh?

Answer 21

can i multiply both equations by that?

Answer 22

so i now have for my numerator(if i am able to multiply both by x^2+xh ) 2x-2x+2h

Answer 23

Ummm I wouldn't multiply out the denominator like that.
It's going to make it harder to see what you actually need in each fraction.

Answer 24

The first fraction has a factor of \(\Large\rm (x+h)\)
while the second fraction has a factor of \(\Large\rm x\)

So our common denominator will be \(\Large\rm x(x+h)\), yes?

Answer 25

oh. ok. i just lost myself

Answer 26

yes

Answer 27

x^2 +xh

Answer 28

Nooo, stop multiplying it >.< lol

Answer 29

Notice that both fractions don't need the entire thing.
The first fraction only needs the \(\Large\rm x\) factor,
and the second fraction only needs the \(\Large\rm (x+h)\) factor.

Answer 30

hahahaha ok sorry i have adhd so i get too excited jhahjaha

Answer 31

yes

Answer 32

so would i have to multiply 2/(x+h) by x ?

Answer 33

and the second equation 2/x by (x+h) ?

Answer 34

or am I going in the completely wrong direction?

Answer 35

Yes good :)
\[\Large\rm \color{royalblue}{\frac{x}{x}}\cdot\frac{2}{(x+h)}\]

Answer 36

\[\Large\rm \color{orangered}{\frac{(x+h)}{(x+h)}}\cdot\frac{2}{x}\]

Answer 37

ok

Answer 38

so now i have [(2x)/(x^2+xh)] - [(2x+2h)/(2+xh) ]

Answer 39

Stop multiplying the bottomssssss >:U
AHHHHHHHHHHHHHHHHHHHHH

Answer 40

:3

Answer 41

oh my goodness i though i was supposed too hahahahaha arent you glad you only have to deal with me tonight ;)

Answer 42

ok so now what?

Answer 43

\[\Large\rm \frac{\dfrac{2}{x+h}-\dfrac{2}{x}}{h}\quad=\quad\frac{\dfrac{2x}{x(x+h)}-\dfrac{2x+2h}{x(x+h)}}{h}\]

Answer 44

oooooooooooooohhhhhhhhhhhhhhhhhhhhhhhh

Answer 45

thats what you wanted me to do :)
ok I promise I am a really smart student, Its just late.

Answer 46

:3

Answer 47

ok back to seriousness.

Answer 48

They have the same denominator, s we can combine the fractions and perform the subtraction if we have any like-terms.

Answer 49

ok

Answer 50

\[\Large\rm \frac{\dfrac{2x-2x-2h}{x(x+h)}}{h}\]

Answer 51

Do you understand why we're subtracting the 2h and not adding it?

Answer 52

yes becuase the 2x+2h is one term and the negative needs to be distributed to both the 2x and 2h

Answer 53

k good :)

Answer 54

ok so I AM still confused on what to do next. and I dont want to jump and try multiplying becuase I am sure I will get it wrong lol

Answer 55

It looks like our numerator has like-terms.
Let's combine the x's.

Answer 56

yes -2h is the only numerator left

Answer 57

\[\Large\rm \frac{\frac{-2h}{x(x+h)}}{h}\]Ok cool.

Answer 58

This next step umm.......
Ooo boy I'm at a loss for words what to call this....

Answer 59

Example:
If I have a half,
\[\Large\rm \frac{1}{2}\]and I divide that by 3,\[\Large\rm \frac{\left(\frac{1}{2}\right)}{3}\]Realize that there is an invisible 1 here,\[\Large\rm \frac{\left(\frac{1}{2}\right)\cdot1}{3}\]So I can rewrite this as:\[\Large\rm \left(\frac{1}{2}\right)\cdot \frac{1}{3}\]

Answer 60

Can* you make sense of that crazy fraction math I just did?
We're going to apply that to our problem.

Answer 61

yes i was actually already trying to do that, but I didnt want to get in trouble "jumping ahead
" :)

Answer 62

\[\Large\rm \frac{\left(\frac{-2h}{x(x+h)}\right)}{h}\]lol :P

Answer 63

{ (-2h) / [x(x+h)] } * [1/h]

Answer 64

do i need to find another common denominator? or do i cross multiply?

Answer 65

you -_-
you people and your cross multiplication..... sigh...

Answer 66

We're multiplying fractions so:

No, we don't need a common denominator,
No, we don't cross multiply.

Answer 67

i dont know wha * 1/h means lol. looks like a multiply the first equation by 1/h

ok so i am wrong,
and i am wrong


daaarrrnnnnn i love math and i though it was so easy, maybe I am just overthinking things

Answer 68

:)

Answer 69

oh so do i just multiply straight across?

Answer 70

Yes :)

Answer 71

yyyyayayayayayayayayaya :)

Answer 72

And again, don't distribute the h to each term in the brackets.

Answer 73

middle school math = so simple :)

Answer 74

Just leave it in front :d

Answer 75

in the front? ok :)

Answer 76

\[\Large\rm \frac{\left(\frac{-2h}{x(x+h)}\right)}{h}=\left(\frac{-2h}{x(x+h)}\right)\cdot \frac{1}{h}\]

Answer 77

yes so now I have (-2)/[x(x+h)] because the h on the bottom and top cancel each other out?

Answer 78

yay :)

Answer 79

So from this point, you can either leave it like that as your final answer (which is what I would do).

Or if you INSIST, then you can multiply out your denominator at this point :d

Answer 80

hahaha ill leave AS IS. no more multiplying, it keeps getting me in trouble lol

Answer 81

thank you. i will ATTEMPT the rest of the problems on my own :)

Answer 82

cool c: gj