Graph the function

Question

Graph the function of g(x)=2+log₃(x+1)

Give its domain and range using interval notation

Answer 1

Erica,

 

Let's look at log₃(x+1) first. Given an x, this function will tell you what power you have to raise 3 to to get x+1.

 

So if x=2, log₃(x+1) = log₃(3) = 1 (because 3 = 3¹)

 

If x = 8, log₃(x+1) = log₃(9) = 2 (because 9 = 3²)

 

If x = 26, log₃(x+1) = log₃(x+1) = 3 (because 27 = 3³)

 

Try plotting those points: (2, 1) ; (8, 2) ; (26, 3).

 

One more point: x=0, log₃(x+1) = log₃(1) = 0 (because 1 = 3⁰)

 

What happens around x=-1? Can you take log³(0)? Another way to ask that is, can 3 be raised to any number such that the result equals zero? No, because 3 raised to anything ≠ 0. So there's either a hole or an asymptote there.

 

What happens to log₃(x+1) when x gets close to -1 from the right? Let's try x=-26/27, log₃(1/27) = -3. (3^-3 = 1/3³ = 1/27).

 

Now x=-80/81, log3(1/81) = -4.

 

Try adding these points: (-26/27, -3) ; (-80/81, -4).

 

So x = -1 is an aysmptote.

 

Does log₃(x+1) exist on the left-hand side of the asymptote? Try x=-100, log₃(-99), which does not exist, since there is no real number one can raise to such that it equals a negative number. So nothing exists on the left-hand side of the asymptote.

 

You should now have the basic shape of log₃(x+1). The only thing left to do is add 2. This will give log₃(x+1) + 2 = 2 + log₃(x+1).

 

Adding 2 will shift the graph up two spaces. Et voila.

 

As for the domain, the graph should make it clear that the doman is (-1, ∞). To find the range, look at how far down the graph goes. All the way to -∞, right? Now how far up does it go? Using your calculator, try plugging in a really large x. Now plug in an even bigger one. Does the function go up? Will it always keep going up? Then what will it be approaching?

 

Best of luck, and if you need any more help, leave me a comment,

Bryce