How do you find the domain and range of #1/(x+6)#?

The domain is all possible values of #x# for which #1/(x+6)# is defined.

We see that the function is only undefined if the denominator is 0, meaning that #x+6 = 0#

This tells us that #x# cannot be #-6#.

So we can say our domain is: #" "x in RR, x ne -6#

(This is just a fancy way of saying "#x# can be all real numbers except for -6")

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Range is a little harder to find. We need to find all possible values that #1/(x+6)# could be.

Let's think about it this way: what does the graph of #y = 1/(x+6)# look like? It will be the graph of #y = 1/x#, but translated 6 units to the left, like this:
graph{y = 1/(x+6) [-13.71, 6.29, -4.76, 5.24]}

We need to find all possible #y# values that can be generated from this graph.

When #x < -6#, we can see that the function #1/(x+6)# will be negative, since #x+6 # will be negative.

As we approach #x = -6# from the left side, the function flies downwards towards #-oo#, hitting every possible negative value.

As we approach #x = -oo#, the function tends towards zero, but never actually reaches it. This is because the denominator is getting bigger and bigger, so the fraction is getting closer and closer to 0 without ever reaching it.

Therefore, from #x = -oo# to #x=-6#, we can say that we will hit all possible negative values of #y#.

The same logic can be used for the positive side of the graph.

As we approach #x = -6# from the right side, the function flies upwards towards #oo#, hitting every possible positive value.

As we approach #x = oo#, the function tends towards zero but never actually reaches it.

Therefore, from #x = -6# to #x = oo#, we can say that we will hit all possible positive values of #y#.

We've checked every possible #x# value now, and our possible #y# values are:

#y# is a real number
#y# is positive OR negative

In other words:

#y in RR, y ne 0#

This is the range of #1/(x+6#

Final Answer