find the difference quotient
f(x) = 4 +3x-x^2
f(3+h) - f(3)/h
please show work

- anonymous

find the difference quotient
f(x) = 4 +3x-x^2
f(3+h) - f(3)/h
please show work

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- anonymous

the whole thing is over h

- amistre64

substitute (3+h) and (3) into your function
then divide by h and use algebraic techniques to solve

- amistre64

you end up with 3 - 2x

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## More answers

- amistre64

if you simply want someone to go thru the effort of typing it all out for you so that you can copy it to your homework; im sure someone will oblige ;)

- anonymous

not true, I am taking a class and I just don't understand how to do some of the problems. I am a single mom going back to school and struggling a bit. Thanks for your help.

- amistre64

then at what part does your understanding begin to fade?

- amistre64

im a single dad going back to college.... so that really doesnt apply does it?

- amistre64

suppose we have the function:
f(x) = x+2
How do we determine: f(5) ??

- anonymous

Good for you apparently you are better off then I am. I am just looking for help ok?

- amistre64

or simply put:
f(x=5) ?

- anonymous

You put 5 in for x

- amistre64

thats right :)
then for these that you have we do that same, but with the x=3 and the x=3+h

- amistre64

what do those look like then?

- anonymous

so I will end up with 2 answers?

- amistre64

youll end up subtracting one function from the other ...yes,
4 +3x-x^2

- amistre64

4 +3(3+h)-(3+h)^2 - [4 +3(x)-(3)^2] = ?

- anonymous

ok I understand now

- amistre64

when all that tedious window dressing is done; you should obtain the result; 3+h-2x for an answer :)

- anonymous

do i multiply by h then to get rid of the h on the bottom?

- amistre64

you can simply divide everything by h to get rid of it; since there is no = sign to equate to, we should avoid doing the multiply by 'h'.

- amistre64

it might be correct, but i dont know how accurate it can be :)

- anonymous

ok

- amistre64

4 + 9 + 3h -9 -6h -h^2 - 4 -9 + 81
------------------------------- is what i get so far, if i didnt mess it up
h

- amistre64

gonna have to work it on paper; i get to many typos this way :)

- amistre64

see, 3^2 = 9; not 81 :)

- anonymous

The following expression replaces the x in the function value by (h+3) subtract the result of replacing the x in the function value by 3 and then dividing this result by h.
\[\left(4+3 x-x^2\text{/.}x\to h+3\right)-\frac{\left(4+3 x-x^2\text{/.}x\to 3\right)}{h} \]= \[9+3 (3+h)-(3+h)^2-3 x \]
Although I could be wrong.

- amistre64

4 + 9 + 3h -9 -6h -h^2 - 4 -9 + 9
-------------------------------
h
-3h -h^2 -3h h^2
-------- = ---- - ---- = -3 -h ... but i gotta recheck myself again :)
h h h

- amistre64

3-2x; as x-> 3 we get
3 -6 = -3; ...its good :)

- amistre64

as h-> 0 we get -3 as an answer, right?

- amistre64

now how can i make this clearer? or are we good with it?

- amistre64

robtob; good effort, wrong interpretation of the expression ;)

- anonymous

amistre64,
Blundered on the first post. I think the following is now correct. At 76 years if age I cannot type with the toes of my left foot and monitor what is going on on the computer screen in real time.
\[\left(4+3 x-x^2\text{/.}x\to h+3\right)-\frac{\left(4+3 x-x^2\text{/.}x\to 3\right)}{h}=4-\frac{4}{h}-3 h-h^2 \]

- amistre64

:)

- amistre64

\[\frac{[4 + 3(3+h) - (3+h)^2] - [4+3(3)-(3)^2]}{h}\]

- amistre64

since the original equation is : 4+3x-x^2...it has to derive down to:
3 -2x; and at f(3) that has to equal -3 when h=0

- amistre64

\[\frac{[4+9-9-4+9-9] -3h - h^2}{h}\]

- amistre64

\[\frac{-3h}{h} - \frac{h^2}{h}\] equals:
-3 - h

- amistre64

when h = 0; this is -3 :)

- anonymous

thank you so much for your help, sorry my internet dropped

- amistre64

'sok; internets a fickle beast :)

- anonymous

amistre64,
googled: difference quotient and looked at the first hit.
http://answers.yahoo.com/question/index?qid=20090320184937AAl0VC6
Got the general idea now.
Thanks.

- anonymous

I hope i am not in over my head going back to school and starting with a calc class. This site has been really helpful.

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