$f(x) = 1/x , [f(x) - f(a)] / (x-a)$
How do I go about setting this up?
If $f(x)= 1/x$ does this term only apply to $f(x)?$
What do I do with the $f(a)? $
Just add $1/a?$
And if so why?
This is confusing if it is the case. I do remember it being so but I do not understand why?
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$\begingroup$i think you are asked to simplify $\dfrac{f(x) - f(a)}{x - a}$ for $f(x) = \dfrac{1}{x}.$
you can do that in the following way: $$ \dfrac{f(x) - f(a)}{x - a} = \frac{\frac{1}{x} - \frac{1}{a}}{x-a} = \dfrac{(a-x)}{ax(x-a)} = -\dfrac{1}{xa}$$
you seem to be confused about how to evaluate or what $f(a)$ means for a given function. the function $f$ is a rule that tells what happens to something usually $x$ when the rule $f$ is applied. this is written as $f(x).$ for example $f(x) = \dfrac{1}{x}$ says that the rule $f$ is to find the reciprocal of what ever is given to it. that is $f(5) = \dfrac{1}{5}, f(-\dfrac{1}{2}) = -2, f(a) = \dfrac{1}{a}$ and so on.
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