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Graph f(x) = log base 3 x-1

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Hello there. To solve this question, we have to remember how to graph a function.

First, remember that the logarithm in any base is an injective function (one-to-one).

We have to find its key-features before graphing it.

These key-features includes: x-intercept, y-intercept (if it exists), its vertical asymptote.

Okay. To find the x-intercept, set the function equal to zero:


f(x)=\log_3(x)-1=0

Adding 1 on both sides, we get


\log_3(x)=1

Apply the property:


\log_a(b)=c\Rightarrow b=a^c

Hence we have


x=3^1=3

So the x-intercept happens at x = 3.

To determine whether or not it has an y-intercept, we check if the limit as x goes to zero from both sides exists and are equal:


\begin{gathered} \lim_(x\to0^+)\log_3(x)-1=-\infty \\ \\ \lim_(x\to0^-)\log_3(x)=-\infty \end{gathered}

In this case, this means we cannot evaluate this function at x = 0, so it doesn't have y-intercept(s).

Next, to determine its vertical asymptote, we check the value of x for which the argument goes to zero, that is


x=0

Therefore it has a vertical asymptote at x = 0.

The graph is then given by:

This is a sketch of the graph of the function.

The final answer to this question is contained in the last option.

Graph f(x) = log base 3 x-1-example-1
User SiberianGuy
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