# Deeper Understanding of Functions in Pre-Calculus

I find the concept of functions to be confusing. The formal definition of a Function is that a function f is a rule that assigns to each element x in a set A exactly one element, called f(x), in set B.

With that definition in mind, let $$f(x)= x+ \frac{2}{x}\cdot$$ Rewrite $$\frac{f(x+h) - f(x)}{h}$$

Is one supposed to input the terms of the difference quotient inside f(x) (function input)? Thus,

$$\frac{x+h+ \frac{2}{x+h}- x+\frac{2}{x}}{h}$$ would be the expression one would need to solve?

Please explain as if I am a 5 year old.

• I edited your post. Please confirm. But note that there is a sign error which I kept. It should be $-\frac{2}{x}\cdot$ With this aside your interpretation is correct. – gammatester Apr 3 '14 at 11:34
• OK. Here the explanation for the missing - sign: $$-f(x) = -\left(x + \frac{2}{x}\right) = -x - \frac{2}{x}$$ – gammatester Apr 3 '14 at 11:40
• I would first teach a 5 year old that an equation is something with an equals sign in it, so there is no equation here, so no equation to solve. What is it you actually want to do with the expression you have written down? – Gerry Myerson Apr 3 '14 at 11:51
• Also, the domain matters, as you said that "a function assigns to each element in a set $A$ exactly one element in a set $B$". To be precise, for what $x$ does $f(x)$ make sense, and $f(x+h)-f(x)/h$? – Luiz Cordeiro Apr 3 '14 at 12:29
• gammaster - Other than that, I have the right idea? – Cetshwayo Apr 3 '14 at 14:01

Note that $f(x+h)=\displaystyle x+h+\frac{2}{x+h}$ and so $$\frac{f(x+h)-f(x)}{h}=\frac{\displaystyle x+h+\frac{2}{x+h}-(x+\frac{2}{x})}{h} =$$ $$=\frac{\displaystyle x+h+\frac{2}{x+h}-x-\frac{2}{x}}{h} =\frac{\displaystyle h+\frac{2}{x+h}-\frac{2}{x}}{h}=$$ $$=\frac{h}{h}+\frac{\displaystyle \frac{2x-2(x+h)}{(x+h)x}}{h}=1+\frac{\displaystyle \frac{2x-2x-2h}{x(x+h)}}{h}=$$ $$=1+\frac{-2h}{x(x+h)}\cdot \frac{1}{h}=1-\frac{2}{x(x+h)}.$$