Complex fraction

[khavar]

Simplify.
{[(x+h)/[(x+h)+2] - 2/(x+2]}/h

{[(x+h)(x+2)]/[(x+h+2)(x+2)]} - {2(x+h+2)/[(x+h+2)(x+2)]}/h =
(x²+hx-4)/[(x+h+2)(x+2)]/h = (x²+hx-4)/[h(x+h+2)(x+2)]

That is as far as I can get. Have been playing with this one since last night. The answer in my book: 2/[(x+h+2)(x+2)], h cannot equal zero.

 

[khavar]

I am not able to paste a Microsoft word equation into this text box, otherwise my expressions would be more cleaned up and more clear. Sorry about that.

 

[JeffM]

Simplify.
{[(x+h)/[(x+h)+2] - 2/(x+2]}/h

{[(x+h)(x+2)]/[(x+h+2)(x+2)]} - {2(x+h+2)/[(x+h+2)(x+2)]}/h =
(x²+hx-4)/[(x+h+2)(x+2)]/h = (x²+hx-4)/[h(x+h+2)(x+2)]

That is as far as I can get. Have been playing with this one since last night. The answer in my book: 2/[(x+h+2)(x+2)], h cannot equal zero.

If you wrote the problem down correctly, your work looks fine to me.

\(\displaystyle \dfrac{\dfrac{x + h}{(x + h) + 2} - \dfrac{2}{x + 2}}{h} = \dfrac{x + h}{h(x + h + 2)} - \dfrac{2}{h(x + 2)} =\)

\(\displaystyle \dfrac{(x + h)(x + 2)}{h(x + h + 2)(x + 2)} + \dfrac{(-2)(x + h + 2)}{h((x + 2)(x + h + 2)} = \dfrac{x^2 + hx + 2x + 2h - 2x - 2h - 4}{h(x + 2)(x + h + 2)} = \dfrac{x^2 + hx - 4}{h(x + 2)(x + h + 2)}.\)

However, I suspect the INTENDED problem was

\(\displaystyle \dfrac{\dfrac{2}{x + h + 2} - \dfrac{2}{x + 2}}{h} = \dfrac{2}{h(x + h + 2)} - \dfrac{2}{h(x + 2)} = \dfrac{2(x + 2)}{h(x + 2)(x + h + 2)} + \dfrac{(-2)(x + h + 2)}{h(x + 2)(x + h + 2)} = \)

\(\displaystyle \dfrac{2x + 4 - 2x - 2h - 4}{h(x + 2)(x + h + 2)} = \dfrac{-2h}{h(x + 2)(x + h + 2)} = \dfrac{-2}{(x + 2)(x + h + 2)}.\)

Note that the answer above is the same as the one indicated in the book except for the minus 2 instead of the plus 2. The above is a little introduction to differential calculus. Perhaps the book misstated the problem. Perhaps you misread the problem.

EDIT: I suspect Soroban made a better guess at the intended problem than I did. Notice that his ended up with the right sign.

 

[soroban]

Hello, khavar!

\(\displaystyle \text{Simplify: }\:\dfrac{\dfrac{x+h}{(x+h)+2} - \dfrac{2}{x+2}}{h}\;\;\color{blue}{\text{ This is incorrect!}}\)

I assume the problem is:

. . \(\displaystyle \boxed{\begin{array}{ccccc}\text{Given:}& f(x) \:=\:\dfrac{x}{x+2} \\ \\ \text{Find:} & \dfrac{f(x+h) - f(x)}{h} \end{array}} \)


\(\displaystyle \dfrac{f(x+h)-f(x)}{h} \;=\;\dfrac{1}{h}\left(\dfrac{x+h}{x+h+2} - \dfrac{x}{x+2}\right)\)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\left(\dfrac{(x+h)(x+2) - x(x+h+2)}{(x+h+2)(x+2)}\right) \)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\left(\dfrac{x^2 + 2x + hx + 2h - x^2 - hx - 2x}{(x+h+2)(x+2)}\right) \)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\cdot\dfrac{2h}{(x+h+2)(x+2)} \)

. . . . . . . . . . . . . \(\displaystyle =\; \dfrac{2}{(x+h+2)(x+2)}\)

 

[khavar]

If you wrote the problem down correctly, your work looks fine to me.

\(\displaystyle \dfrac{\dfrac{x + h}{(x + h) + 2} - \dfrac{2}{x + 2}}{h} = \dfrac{x + h}{h(x + h + 2)} - \dfrac{2}{h(x + 2)} =\)

\(\displaystyle \dfrac{(x + h)(x + 2)}{h(x + h + 2)(x + 2)} + \dfrac{(-2)(x + h + 2)}{h((x + 2)(x + h + 2)} = \dfrac{x^2 + hx + 2x + 2h - 2x - 2h - 4}{h(x + 2)(x + h + 2)} = \dfrac{x^2 + hx - 4}{h(x + 2)(x + h + 2)}.\)

However, I suspect the INTENDED problem was

\(\displaystyle \dfrac{\dfrac{2}{x + h + 2} - \dfrac{2}{x + 2}}{h} = \dfrac{2}{h(x + h + 2)} - \dfrac{2}{h(x + 2)} = \dfrac{2(x + 2)}{h(x + 2)(x + h + 2)} + \dfrac{(-2)(x + h + 2)}{h(x + 2)(x + h + 2)} = \)

\(\displaystyle \dfrac{2x + 4 - 2x - 2h - 4}{h(x + 2)(x + h + 2)} = \dfrac{-2h}{h(x + 2)(x + h + 2)} = \dfrac{-2}{(x + 2)(x + h + 2)}.\)

Note that the answer above is the same as the one indicated in the book except for the minus 2 instead of the plus 2. The above is a little introduction to differential calculus. Perhaps the book misstated the problem. Perhaps you misread the problem.

EDIT: I suspect Soroban made a better guess at the intended problem than I did. Notice that his ended up with the right sign.

Thank you once again, JeffM. This is one of several typographical errors in my math book. One thing I will say is that the errors make me second guess my work and I come out learning the particular process even better. Thankfully this forum exists.

 

[khavar]

Hello, khavar!


I assume the problem is:

. . \(\displaystyle \boxed{\begin{array}{ccccc}\text{Given:}& f(x) \:=\:\dfrac{x}{x+2} \\ \\ \text{Find:} & \dfrac{f(x+h) - f(x)}{h} \end{array}} \)


\(\displaystyle \dfrac{f(x+h)-f(x)}{h} \;=\;\dfrac{1}{h}\left(\dfrac{x+h}{x+h+2} - \dfrac{x}{x+2}\right)\)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\left(\dfrac{(x+h)(x+2) - x(x+h+2)}{(x+h+2)(x+2)}\right) \)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\left(\dfrac{x^2 + 2x + hx + 2h - x^2 - hx - 2x}{(x+h+2)(x+2)}\right) \)

. . . . . . . . . . . . . \(\displaystyle =\;\dfrac{1}{h}\cdot\dfrac{2h}{(x+h+2)(x+2)} \)

. . . . . . . . . . . . . \(\displaystyle =\; \dfrac{2}{(x+h+2)(x+2)}\)

Thank you, Soroban. I appreciate the time that was taken to assist me.