Complex fraction

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khavar

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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.
 
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.
 
khavar said:
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.
Click to expand...
If you wrote the problem down correctly, your work looks fine to me.

x+h(x+h)+2−2x+2h=x+hh(x+h+2)−2h(x+2)=\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)} =

(x+h)(x+2)h(x+h+2)(x+2)+(−2)(x+h+2)h((x+2)(x+h+2)=x2+hx+2x+2h−2x−2h−4h(x+2)(x+h+2)=x2+hx−4h(x+2)(x+h+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

2x+h+2−2x+2h=2h(x+h+2)−2h(x+2)=2(x+2)h(x+2)(x+h+2)+(−2)(x+h+2)h(x+2)(x+h+2)=\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)} =

2x+4−2x−2h−4h(x+2)(x+h+2)=−2hh(x+2)(x+h+2)=−2(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.
 
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Hello, khavar!

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

I assume the problem is:

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


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

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

. . . . . . . . . . . . . =  1h(x2+2x+hx+2h−x2−hx−2x(x+h+2)(x+2))\displaystyle =\;\dfrac{1}{h}\left(\dfrac{x^2 + 2x + hx + 2h - x^2 - hx - 2x}{(x+h+2)(x+2)}\right)

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

. . . . . . . . . . . . . =  2(x+h+2)(x+2)\displaystyle =\; \dfrac{2}{(x+h+2)(x+2)}
 
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JeffM said:
If you wrote the problem down correctly, your work looks fine to me.

x+h(x+h)+2−2x+2h=x+hh(x+h+2)−2h(x+2)=\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)} =

(x+h)(x+2)h(x+h+2)(x+2)+(−2)(x+h+2)h((x+2)(x+h+2)=x2+hx+2x+2h−2x−2h−4h(x+2)(x+h+2)=x2+hx−4h(x+2)(x+h+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

2x+h+2−2x+2h=2h(x+h+2)−2h(x+2)=2(x+2)h(x+2)(x+h+2)+(−2)(x+h+2)h(x+2)(x+h+2)=\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)} =

2x+4−2x−2h−4h(x+2)(x+h+2)=−2hh(x+2)(x+h+2)=−2(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.
Click to expand...

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.
 
soroban said:
Hello, khavar!


I assume the problem is:

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


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

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

. . . . . . . . . . . . . =  1h(x2+2x+hx+2h−x2−hx−2x(x+h+2)(x+2))\displaystyle =\;\dfrac{1}{h}\left(\dfrac{x^2 + 2x + hx + 2h - x^2 - hx - 2x}{(x+h+2)(x+2)}\right)

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

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

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