[f(x+deltax)-f(x)]/delta x

**yess** · September 23rd 2010, 08:37 AM

algebraically simplify [f(x+deltax)-f(x)]/delta x to -1/[(x+delta x-3)(x-3)] for f(x) = (x-2)/(x-3)

**Soroban** · September 23rd 2010, 12:58 PM

Hello, yess!

$\text{Given: }\;f(x) \:=\:\dfrac{x-2}{x-3}$

$\text{Simply }\,\dfrac{f(x+\Delta x) - f(x)}{\Delta x}\:\text{ to }\:\dfrac{-1}{(x+\Delta x - 3)(x - 3)}$

When working with a Difference Quotient, I make three steps out of it. .**

. . [1] Find $f(x+\Delta x)$ . . . Replace $x$ with $x + \Delta x$ ... and simplify.

. . [2] Subtract $f(x)$ . . . Subtract the original function ... and simplify.

. . [3] Divide by $\Delta x$ . . . Write $\Delta x$ in the denominator ... and reduce.

Here we go . . .

$[1]\;\;f(x+\Delta x) \;=\;\dfrac{(x+\Delta x) - 2}{(x+\Delta x) - 3}$

$[2]\;\;f(x+\Delta x) - f(x) \;=\;\dfrac{x+\Delta x - 2}{x + \Delta x - 3} - \dfrac{x-2}{x -3}$

. . $=\;\dfrac{(x-3)(x+\Delta x - 2) - (x-2)(x+\Delta x - 3)}{(x+\Delta x - 3)(x - 3)}$

. . $=\;\dfrac{x^2 + x\,\Delta x - 2x - 3x - 3\Delta x + 6 - x^2 - x\,\Delta x + 3x + 2x + 2\Delta x - 6}{(x+\Delta x - 3)(x - 3)}$

. . $=\;\dfrac{-\Delta x}{(x+\Delta x - 3)(x - 3)}$

$[3]\;\;\dfrac{f(x+\Delta x) - f(x)}{\Delta x} \;=\;\dfrac{-\Delta x}{\Delta x\, (x + \Delta x - 3)(x - 3)}$

. . . . . . . . . . . . . . . . . $=\;\dfrac{-1}{(x+\Delta x - 3)(x - 3)}\quad\hdots\;There!$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Am I the only teacher on the entire planet who taught it this way?

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