Question

PLEASE HELP: College Algebra

Answer 1

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

Answer 2

Try adding the fractions in the numerator (common denominator)

Answer 3

(I will use Q for this quotient.)

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}}\)

Answer 4

so first I have to find the GCf?

Answer 5

No, they don't have a GCF

Answer 6

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{1\color{red}{\times x^2}}{(x+h)^2\color{red}{\times x^2}}-\frac{1\color{red}{\times (x+h)^2}}{x^2\color{red}{\times (x+h)^2}}}{h}}\)

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{x^2}{x^2(x+h)^2}-\frac{(x+h)^2}{x^2(x+h)^2}}{h}}\)

Answer 7

tell me if you see what I did and why I did so (ok?)

Answer 8

That's what I meant to do, i got to that point... you made it have a common denom so that subtracting would be possible.

Answer 9

yep ^^

Answer 10

perhaps, fibo .... the product of the two denoms is the LCD, tho' :)

Answer 11

Would be easier if it was h->0

Answer 12

anyway ...

Answer 13

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}}\)

then I combined the terms

Answer 14

Simplify (expand and subtract/add like-terms) ... and you will get a nice result.

Answer 15

Can you expand and simplify (?):

\(x^2-(x+h)^2=\)

Answer 16

don't eh x+h's cancel?

Answer 17

where? Before subtracting the fractions?

Answer 18

okay so I expanded to get \[h^2+x^2+2hx\]

Answer 19

Yes, \((x+h)^2=x^2+h^2+2hx\).

Answer 20

So,

\(x^2-(x+h)^2=?\)

Answer 21

\[x^2-x^2-h^2-hx\]

Answer 22

yes, very good, and that simplifies to (?) ...

Answer 23

\[-h^2-hx\]

Answer 24

Yups

Answer 25

So, we had:

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}}\)

and, as we have worked out, it turns into

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{-h^2-hx}{x^2(x+h)^2}}{h}}\)

Answer 26

You can divide by \(h\) on top and bottom.

Answer 27

((This will give you just a single fraction, as opposed to a nested fraction.))

Answer 28

how?

Answer 29

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{-h^2-hx}{x^2(x+h)^2}\color{red}{\div h}}{h\color{red}{\div h}}}\)

Answer 30

ooh okay

Answer 31

Alright .... tell me what you then get ((take you time))

Answer 32

If you want, you can factor the \(-h^2-hx\) out of \(h\), first
(before dividing by \(h\) on top and bottom)

Answer 33

Oh, hold !

Answer 34

The expansion is wrong

Answer 35

\(\color{blue}{\displaystyle (x+h)^2=x^2+h^2+2hx}\)

Answer 36

So,

\(x^2-(x+h)^2=-h^2-2hx\)

Answer 37

So, we actually have:

\(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{-h^2-2hx}{x^2(x+h)^2}}{h}}\)

Answer 38

alright , so far?

Answer 39

yes

Answer 40

Can you factor \(-h^2-2hx\) out of \(h\) ?

Answer 41

can't I factor it out of -h?

Answer 42

Yes, you can do that as well :)

Answer 43

-h(1^2+2x)

Answer 44

\(-h^2=-h\times h\)
\(-2hx=-h\times 2x\)

\(-h^2-2hx=-h(h+2x)\)

Answer 45

So, \(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{-h^2-2hx}{x^2(x+h)^2}}{h}}\), is equivalent of this, \(\color{blue}{\displaystyle Q=\frac{\displaystyle \frac{-h(h+2x)}{x^2(x+h)^2}}{h}}\).

Answer 46

now, divide by \(h\) on top and bottom

Answer 47

\(\color{blue}{\displaystyle Q=\frac{\displaystyle ~~~~\frac{-h(h+2x)}{x^2(x+h)^2}\color{red}{\div h}~~~~}{h\color{red}{\div h}}}\)

Answer 48

\[\frac{ -h-2hx }{ hx^2(hx+h^2)^2 }\]

Answer 49

your denominator is off, not sure exactly what you did, but please check your algebra

Answer 50

\(\color{blue}{\displaystyle Q=\frac{\displaystyle ~~~~\frac{-h(h+2x)}{x^2(x+h)^2}\color{red}{\div h}~~~~}{h\color{red}{\div h}}=\frac{\displaystyle ~~~~\frac{-\cancel{h}(h+2x)}{x^2(x+h)^2}\color{red}{\div \cancel{h}}~~~~}{\bcancel{h}\color{red}{\div \bcancel{h}}}=\frac{\displaystyle \frac{-(h+2x)}{x^2(x+h)^2}}{1}}\)

Answer 51

\(\color{blue}{\displaystyle Q=\frac{-h-2x}{x^2(x+h)^2}}\)

Answer 52

If anything doesn't make sense, then ask ...

Answer 53

****************
This is just a check for me (and for you, if you've done calculus I)

\(\color{blue}{\displaystyle \lim_{h\to 0}\frac{\displaystyle \frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}=\frac{d}{dx}(x^{-2})=-2x^{-3}=\frac{-2}{x^3}}\)

\(\color{blue}{\displaystyle \lim_{h\to 0}\frac{-h-2x}{x^2(x+h)^2}=\frac{-2x}{x^2(x)^2}=\frac{-2x}{x^4}==\frac{-2}{x^3}=\frac{d}{dx}(x^{-2})}\)

Answer 54

Oh crap ... you can't see it ?

Answer 55

It worked :D

Answer 56

Oh, nice :)

Answer 57

Don't worry about my calculus check (if you aren't math major, you probably wouldn't ever need calculus classes)

Answer 58

I will NEVER major in math :P. I'm still in HS anyway .-.

Answer 59

it makes sense though :D thank you

Answer 60

Not a problem ... yw