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How to draw a graph $f(x)=-\log_{3}(3(x-1))$?

What's with that 3 before x, should I expose it? so I get $f(x)=-\log_{3}(3(x-1))$?

So how do I then draw it? The process (of how I would draw this function):

1.)$f(x)=\log_{3}x$

2.)$f(x)=\log_{3}3x$

3.)$f(x)=\log_{3}(3(x-1))$

4.)$f(x)=-\log_{3}(3(x-1))$

Josephine Moeller
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user114141
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    One thing that will make this easier is to use $\log$ identities to pull the $3$ out: $-\log_3(3(x-1))$ $\ =\ $ $-\log_3(3)-\log_3(x-1)$ $\ =\ $ $-1-\log_3(x-1)$. – Josephine Moeller Dec 17 '13 at 20:21
  • Thank you! That sure is much easier to draw! :) – user114141 Dec 17 '13 at 20:23
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    No problem. A rule of thumb to remember is that a constant factor inside the function just stretches or shrinks in the $x$ direction. Constants greater than 1 shrink the graph, while constants less than one stretch the graph. Negative constants inside the function flip the x axis. – Josephine Moeller Dec 17 '13 at 20:33

3 Answers3

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You may want to use the following:

  • There is a vertical asymptote when the argument of the $\log$ is zero ($x = 1$).
  • The function crosses zero when the argument of the $\log$ is one ($x = 4/3$).
  • For large enough $x$, the function behaves like $-\log_3(3x) = -1-\log_3(x)$
Nathaniel Bubis
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Usually we draw this kind of elementary functions by mathematical softwares. On the other hand, your process is correct and feasible. It helps understand the rough figure of $f$.

kwgl
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  • But my graph does not match with one in the solutions. If you look at $f(x)=log_{3}3x$,what exactly should I do right here? Should I multipy every x of y? – user114141 Dec 17 '13 at 20:21
  • I do not quite understand what you want to do. There is no standard way of drawing by hand. If you really need to draw by hand, usually we simplify it first, like John suggested, then to figure out its defining domain, then to locate the key points, like Nathaniel said, and then draw the figure according to its continuity. I do not kwon what your newly mentioned $y$ is. – kwgl Dec 17 '13 at 20:34
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Let me give you a summary of what you should do. Firstly you should know that the logarithm function is defined only for positive values. Therefore $3(x-1)>0\rightarrow x>1$. When $f(x)=0$ we have, $3x-3=1\rightarrow x=\frac{4}{3}$. Now we shall consider the derivative of this function.

$$f'(x)=-\frac{1}{(x-1)\ln 3}$$

$f'(x)<0$ for all $x>1$.

We also see that, $x\rightarrow \infty\Rightarrow f(x)\rightarrow-\infty$ and when $x\rightarrow 1^+\Rightarrow f'(x)\rightarrow\infty$.

With all the above information we can draw the graph of $f$ and it turns out like >>this<<.