+216 Let
f(x)=11+21+3x.
There are three real numbers x that are not in the domain of f(x). What is the sum of those three numbers?
+130466 obviously....x cannot = 0
1 + 3/x cannot = 0 ...so 1 + 3/x = [ x + 3] / x
And 2/ ( [ x + 3] / x) = 2x / [ x + 3]
So x cannot = -3
Lastly
1 + [2x] / [ x + 3] cannot = 0
[ x + 3 + 2x] /[x + 3]
[ 3x + 3 ] / [x + 3]
So...... 1 / [ (3x + 3) / (x + 3) ]
(x + 3 ) / (3x + 3)
So x cannot = -1/3
