Ask your own question, for FREE!
Mathematics 51 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

the whole thing is over h

OpenStudy (amistre64):

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

OpenStudy (amistre64):

you end up with 3 - 2x

OpenStudy (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 ;)

OpenStudy (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.

OpenStudy (amistre64):

then at what part does your understanding begin to fade?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (amistre64):

or simply put: f(x=5) ?

OpenStudy (anonymous):

You put 5 in for x

OpenStudy (amistre64):

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

OpenStudy (amistre64):

what do those look like then?

OpenStudy (anonymous):

so I will end up with 2 answers?

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (anonymous):

ok I understand now

OpenStudy (amistre64):

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

OpenStudy (anonymous):

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

OpenStudy (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'.

OpenStudy (amistre64):

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

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (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.

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (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. \[\left(4+3 x-x^2\text{/.}x\to h+3\right)-\frac{\left(4+3 x-x^2\text{/.}x\to 3\right)}{h}=4-\frac{4}{h}-3 h-h^2 \]

OpenStudy (amistre64):

:)

OpenStudy (amistre64):

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

OpenStudy (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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

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

OpenStudy (amistre64):

when h = 0; this is -3 :)

OpenStudy (anonymous):

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

OpenStudy (amistre64):

'sok; internets a fickle beast :)

OpenStudy (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.

OpenStudy (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.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!