How do you do problems such as this?

Evaluate $\displaystyle \frac{f(x) - f(2)}{x - 2}$ for $\displaystyle f(x) = $ $\displaystyle \frac{4}{x}$

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- Jan 26th 2011, 01:59 PMepimetheusaverage rate of change - evaluating functions
How do you do problems such as this?

Evaluate $\displaystyle \frac{f(x) - f(2)}{x - 2}$ for $\displaystyle f(x) = $ $\displaystyle \frac{4}{x}$ - Jan 26th 2011, 02:28 PMAryth
Plug and chug:

$\displaystyle \frac{\frac{4}{x} - 2}{x - 2} = \frac{\frac{4 - 2x}{x}}{x-2} = \frac{\frac{2(2 - x)}{x}}{x-2} = \frac{-2}{x}$ - Jan 26th 2011, 02:43 PMepimetheus
Is it the same thing when you have something like $\displaystyle \frac{f(3 + h) - f(3)}{h}$ for x² - 4?

- Jan 26th 2011, 02:50 PMAryth
Yes, it's exactly the same:

$\displaystyle \frac{(3 + h)^2 - 4 - (9 - 4)}{h} = \frac{9 + 6h + h^2 - 4 - 9 + 4}{h} = \frac{6h + h^2}{h} = 6 + h$