f(x)=3x-8
g(x)=3/(x-10)
find the domain of f+g
The domain of a function is : all possible x that we can 'plug' in our formula with a result that is a real number (not infinity or square root of a negative number or non-positive logarithm).
In usual functions, all x work fine. Usually, the problems araise when you have your x in the denominator of a fraction or inside a root (even ones only : square root, forth root...) or when having your x inside a logarithm.
1) f(x)=3x-8. This function has no problem. I can put any real number x I want and it gives me a value for f(x). Ex. : x=4.5 gives f(4.5) = 3 (4.5) - 8 = 5.5.
So any real number x works. So the domain is all real numbers. We can write dom(f) = R. Where dom(f) means domain of f and R the set of real numbers.
2) g(x)=3/(x-10). We see the x in the denominator of a fraction. This tells us that there might be a problem. With what x ? The formula 3/(x-10) craches when x-10 = 0 (we would have 3/0 which is infinity). We know that x-10=0 if and only if x=10. So x=10 is our only problem maker here.
So all real numbers x except 10 work and give us a result g(x). So the domain of g(x) : dom(g)=(R without 10) =R\{10}. This is just a notation meaning that we remove 10 from the set of real numbers to get the domain of g.
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Now, we add f and g
(f+g)(x) = 3x - 8 + 3/(x-10). Here we sum so we do not have a new fraction or a square root or a logarithm appearing due to our operation of addition between f and g so no new worries. We just take care of our old worries :
-) no worries for f
-) one worry in g : the 10.
So f+g will not give a resulat for all the x which do not work in f and all those which do not work in g. So in total here we just have 10 which is a problem. So dom(f+g) = (R without 10) = R\{10}.
If you know intersection of sets I did here intersection of domain of f and domain of g. Because sum does not add more trouble.
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Here is another example : h(x) = 3/x and m(x) = 3/(x-1).
dom(h) = R\{0} (we take all real numbers except 0 because h(0)=3/0=infinity) and
dom(m)=R\{1} for the same reason.
So h+m = 3/x + 3/(x-1). Both h and m give their problems. So the domain of h+m is R\{0,1} so it is all real numbers except 0 and 1 because (h+m)(0) craches in 3/x and (h+m)(1) craches in 3/(x-1) so craching in one place craches the whole equation.