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anonymous
 5 years ago
find the difference quotient of f; that is, find
(f(x+h)f(x))/h f(x)=x^2+5X1
anonymous
 5 years ago
find the difference quotient of f; that is, find (f(x+h)f(x))/h f(x)=x^2+5X1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sub in the (x+h) for the values of x in f(x), and subtract f(x). Expand out the f(x+h) and simplify all of your variables. You should end up with\[(h^2+2xh+5h)/h\], the h factors out and you 2x+5, if you take the limit as h goes to zero, which is actually what you are trying to get at in most general purposes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so it would be f(x^2+5x1 +h)X^2+5x1 divide by h

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually it is \[([(x+h)^2 +5(x+h)1]  [x^2+5x1]) / h\]. You see, for every value of x in f(x), we are putting x+h in. It is just like putting a constant in like f(5), only it is f(x+h). Does that make sense?. Then you go ahead and expand everything out.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im not sure how to expand it =(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay, the next step is \[([x^2+2xh+h^2+5x+5h1]x^25x+1)/h\]. You go ahead and square the x + h, and distribute the 5 to (x+h), and make sure to distribute the negative to the f(x) values.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have any problems factoring or multiplying variables and constants, like (x+5)(x2), or anything like that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wats the step after that cuz i got that part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(2xh+5h+h^2)/h , that turns into 2x + 5 +h after factoring out the h. .If you were to take the limit as I mentioned, you would sub in 0 for h and get 2x+5. Work on getting the factoring and multiplying down, because it's pretty common where you're at right now and onward. If you have any other questions feel free to ask. : )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No problem, thanks for the nod
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