addition of ordinates:
y = f(x) + g(x)
sketch f(x) and g(x). run vertical lines down the page (lines should hit f(x) and g(x)). add the y-values to construct req graph. sometimes f(x) and g(x) have restricted domains, in which case the vertical lines you run down the page will not hit both f(x) and g(x). obviously you can't add one y-value to something that doesn't exist. this is why the domain of the new graph is given by the intersection of the domains of f(x) and g(x).
product of two functions:
y = f(x)*g(x)
same principle as addition of ordinates. draw f(x) and g(x), run vertical lines down the page, but this time MULTIPLY the y-values. a bit annoying, but still manageable. i've never seen an exam question that asks you to draw the product of two functions from 'scratch'.
composite functions:
y=f(g(x))=fog(x)
it is simply too hard to draw composite functions using the technique described above (i.e. by first drawing f(x) and g(x) and then trying to work out, for each x-value, what the corresponding y-value of the new functions will be). however you do need to know what composite functions are.
the best way to think of composite functions is by visualising g-machines and f-machines. so visualising a production line. in this case, the g-machine is in front of f-machine. now say you have a stack of x-values (literally imagine you have at hand all real numbers). now stuff these x-values one by one into the g-machine. but the g-machine is very selective and will not take certain x-values. say the g-machine doesn't like x-values less than 0. so every time you stuff an x-value less than 0 into the g-machine, it will reject it. the set of all x-values that the g-machine does accept is called the DOMAIN of g. now each time you input an acceptable x-value into the g-machine, it will spit out a new value according to some algorithm/rule. let us call this a g(x)-value. so you've tried every single x-value in the stack you have at hand and fed it into the g-machine. it rejected some, and accepted some. those which the g-machine accepted got converted into g(x)-values. the set of all g(x)-values is called the RANGE of g(x). now you take all the g(x)-values that the g-machine spat out and you feed it into the f-machine, which will convert them into f(g(x)) values according to another algorithm/rule. the set of all f(g(x)) values is termed the RANGE of f(g(x)). now here's the thing:the f-machine is, like the g-machine, really picky; and if the f-machine rejects ANY of the g(x)-values you feed into it, the whole process will stuff up, because no f(g(x)) value will outputted. so in order for f(g(x)) to be DEFINED, the range of g (the set of all g-values outputted by the g-machine) must be a SUBSET of the domain of f (ALL the numbers which the f-machine accepts). if this is not the case, then the function is UNDEFINED. if this IS the case, then we also know that the domain of f(g(x)) (all the x-values that are ultimately accepted and converted into f(g(x)) values) is EQUAL to the domain of g(x), since x-values which the g-machine accepted would have made it to the end, and been converted into an f(g(x)) value. whenever you are confronted with a composite function question, think this.