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$f(x) = 1/x , [f(x) - f(a)] / (x-a)$

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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  • $\begingroup$ For all no-zero numbers $y$, $f(y)=\frac 1 y$. This is true with $y=1,y=2,y=\pi, y=x, y=a,$... $\endgroup$
    – Git Gud
    Feb 14, 2015 at 15:40

1 Answer 1

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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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  • $\begingroup$ why are you allowed to make replace 1/x with 1/a if it states f(x) = 1/x? $\endgroup$
    – Cetshwayo
    Feb 14, 2015 at 15:42
  • $\begingroup$ If $f(x) = 1/x$ then $f(a) = 1/a$... $\endgroup$
    – Eff
    Feb 14, 2015 at 15:43
  • $\begingroup$ @Cetshwayo, i was editing my answer while you were commenting. see if the addition helps you with the notion of function. $\endgroup$
    – abel
    Feb 14, 2015 at 15:47
  • $\begingroup$ Great explanation Abel. $\endgroup$
    – Cetshwayo
    Feb 15, 2015 at 17:45

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