If f(x) = 2/x, simplify [f(x + delta-x) - f(x)] / delta-x

[cherica123]

I don't really know how to do this problem at all, my teacher hinted that it was an into to derivatives or something:

If f(x)= 2/x, simplify:

. . .\(\displaystyle \L \frac{f(x\,+\,\Delta x)\,-\,f(x)}{\Delta x}\)

Obviously I'd plus (2/x) in for the f(x)'s, but im not sure what the delta-x does
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Edited by stapel -- Reason for edit: formatting

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
f(x) & = & \frac{2}{x} \\
\frac{{f(x + \Delta x) - f(x)}}{{\Delta x}} & = & \frac{{\frac{2}{{x + \Delta x}} - \frac{2}{x}}}{{\Delta x}} \\
& = & \frac{{2x - 2\left( {x + \Delta x} \right)}}{{\Delta x\left( x \right)\left( {x + \Delta x} \right)}} \\
& = & \frac{{ - 2\Delta x}}{{\Delta x\left( x \right)\left( {x + \Delta x} \right)}} \\
& = & \frac{{ - 2}}{{\left( x \right)\left( {x + \Delta x} \right)}} \\
\end{array}\)

 

[stapel]

<alert="pet peeve">Yes, this relates to derivatives, a topic you'll encounter in calculus. No, you probably won't remember this by then.</alert>

Don't worry about what delta-x might "mean"; that's another issue for another day. For now, just work with the simplification. And, with simplicity in mind, let's replace "\(\displaystyle \Delta x\)" with "h". And let's work in smaller, more manageable steps:

. . . . .f(x) = 2/x

. . . . .f(x + h) = 2/(x + h)

. . . . .f(x + h) - f(x) = 2/(x + h) - 2/x

Convert to a common denominator:

. . . . .[2(x)] / [x(x + h)] - [2(x + h)] / [x(x + h)]

. . . . .[2x] / [x(x + h)] - [2x + 2h] / [x(x + h)]

. . . . .[2x - 2x - 2h] / [x(x + h)]

. . . . .[-2h] / [x(x + h)]

Now put that (simplified) numerator over the denominator "h", and cancel.

(You can convert back from "h" to "\(\displaystyle \Delta x\)" for the final answer.)

Hope that helps a bit.

Eliz.