Question

Let f(x) = 1/sqrt(x+1) , find (f(x+h)-f(x))/h .
Simplify as much as possible.

Answer 1

lots of ugly algebra
so lets go slow and take it step by step
first off, what is \(f(x+h)\) in this case ?

Answer 2

Would it be \[(1/\sqrt{x+1})+h ?\]

Answer 3

no

Answer 4

that is not \(f(x+h)\) that is \(f(x)+h\)

Answer 5

where you see an \(x\) in \(f(x)=\frac{1}{\sqrt{x+1}}\) replace it by \(x+h\)

Answer 6

Sorry, I'm kind of confused! Algebra is not my forte. Would the equation then be \[1/ \sqrt{x+h+1} ?\]

Answer 7

yes good

Answer 8

now if you don't like algebra, you are really not going to like the next part

Answer 9

we have
\[f(x+h)=\frac{1}{\sqrt{x+h+1}}\] so
\[f(x+h)-f(x)=\frac{1}{\sqrt{x+h+1}}-\frac{1}{\sqrt{x+1}}\] and now you have to actually do the subtraction
any ideas?

Answer 10

Convert the two fractions to common denominators?

Answer 11

by rationalizing the denominators

Answer 12

yeah, but don't do too much work
leave the denominator in factored form
\[\frac{1}{\sqrt{x+h+1}}-\frac{1}{\sqrt{x+1}}=\frac{\sqrt{x+1}-\sqrt{x+h+1}}{\sqrt{x+h+1}\sqrt{x+1}}\]

Answer 13

that wasn't too bad, that is just like \(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}\)

Answer 14

now comes the real kicker
rationalize the NUMERATOR
don't multiply anything out in the denominator at all, leave it in factored form
but multiply top and bottom by \(\sqrt{x+h+1}+\sqrt{x+1}\)

Answer 15

ok that was a mistake, the conjugate of the numerator is
\[\sqrt{x+1}+\sqrt{x+h+1}\]

Answer 16

still with me or did i lose you yet?

Answer 17

Kiiiiinda but I'm kind of confused as of how you'd subtract the \[\sqrt{x+h+1}\]

Answer 18

messed that up
\[\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}\]

Answer 19

so i didn't really "subtract" i just wrote it

Answer 20

\[ \it's \sqrt{x+1}-\sqrt{x+h+1}/\sqrt{x+1}\sqrt{x+h+1}\]

Answer 21

\[\frac{1}{\sqrt{x+h+1}}-\frac{1}{\sqrt{x+1}}=\frac{\sqrt{x+1}-\sqrt{x+h+1}}{\sqrt{x+h+1}\sqrt{x+1}}\] is just subtraction
notice we did not multiply anything out or combine like terms or any algebra at all,

Answer 22

Is that the final answer? i don't thinkyou can simplify that any more

Answer 23

oh not even close

Answer 24

\[\frac{\sqrt{x+1}-\sqrt{x+h+1}}{\sqrt{x+h+1}\sqrt{x+1}}\times \frac{\sqrt{x+1}+\sqrt{x+h+1}}{\sqrt{x+1}+\sqrt{x+h+1}}\]\]

Answer 25

leave the denominator in factored form, but a miracle occurs in the numerator

Answer 26

this is the rationalizing the numerator part

Answer 27

\(a-b)(a+b)=a^2-b^2\) so the numerator gives
\[x+1-(x+h+1)=-h\]

Answer 28

the \[\sqrt{x+1}\] cancels out right?

Answer 29

Oh, I see!

Answer 30

yeah both square roots go
you have to only be careful with the subtraction, make sure you realize the numerator is \(-h\) and not \(h\)
only one more step...

Answer 31

Yay! Getting close :) So at this point, the equation is \[ \frac{{-h} }{\sqrt{x+1}+\sqrt{x+h+1} } \] isn't it?

Answer 32

you are missing a huge part of the denominator

Answer 33

\[\frac{\sqrt{x+1}-\sqrt{x+h+1}}{\sqrt{x+h+1}\sqrt{x+1}}\times \frac{\sqrt{x+1}+\sqrt{x+h+1}}{\sqrt{x+1}+\sqrt{x+h+1}}\]
\[=\frac{-h}{(\sqrt{x+h+1}\sqrt{x+1})(\sqrt{x+1}+\sqrt{x+h+1}}\]

Answer 34

don't even think about multiplying out in the denominator, it will be a huge mess
just leave it

Answer 35

That's the final answer?

Answer 36

no

Answer 37

remember that \(h\) in the denominator we have been ignoring this whole time?
\[\frac{f(x+h)-f(x)}{\color{red}h}\]

Answer 38

it cancels with the \(-h\) in the numerator, leaving a \(-1\) up top

Answer 39

@satellite73 can you help alittle more when you're done ?

Answer 40

\[\frac{-1}{(\sqrt{x+h+1}\sqrt{x+1})(\sqrt{x+1}+\sqrt{x+h+1}}\] is the final answer

Answer 41

Thanks! But, wouldn't the numerator be -1-\[\frac{ 1 }{ \sqrt{x+1} }\] ? because of the extra -f(x) after the f(x+h)? Or did we already do that and I just forgot?

Answer 42

numerator is \(-1\) for sure, because before cancelling it was \(-h\)
the step to making it \(-h\) was \[(\sqrt{x+1}-\sqrt{x+h+1})(\sqrt{x+1}+\sqrt{x+h+1})=x+1-(x+h+1)=-h\]

Answer 43

Oh, alright! I see now. Thanks for you help and patience. :) Sorry, that was for a summer packet for precal and they wanted us to figure how to do these problems by ourselves. Thanks again!

Answer 44

yw
this was not that enlightening, just a lot of algebra
in calc you will repeat this process, then replace \(h\) by \(0\) to get the "derivative"
fortunately there are lots of shortcut methods so you wont have to do this much