Expert: Paul Klarreich - 1/5/2008

Question
Hey,

I am a first year calculus student. We are currently on the chapter with logarithmic functions and their derivatives. I am having problems with answering this question.

We are to determine the relation of the following functions to the graph of y = ln x .

We have to relate it to the domain, range , x - intercept, vertical asymptote etc. Any help will be appreciated.

I know that the graph of y = ln x has a domain from 0 to positive infiniti. I also know that its range is from - infiniti to + infiniti. It has a vertical asymptote at y = 0 and a x-intercept at x = 1. Now we have to relate the following functions to the graph details of ln x.

Here are the functions.

y = 3 ln x

y = -2 ln x

y = ln 2x

y = ln absolute value of x

y = ln (x-1)

y = 2 + ln x

Please help me with these problems as I am finding it real difficult to get the answer. Thanks for all your help and concern.

Regards
Salim Khan

Answer
Questioner:   salim khan
Category:  Calculus
Private:  No
 
Subject:  Calculus first year question
Question:  Hey,

I am a first year calculus student. We are currently on the chapter with logarithmic functions and their derivatives. I am having problems with answering this question.

We are to determine the relation of the following functions to the graph of y = ln x .

We have to relate it to the domain, range , x - intercept, vertical asymptote etc. Any help will be appreciated.

I know that the graph of y = ln x has a domain from 0 to positive infiniti. I also know that its range is from - infiniti to + infiniti. It has a vertical asymptote at y = 0

>> No, you mean x = 0.

and a x-intercept at x = 1. Now we have to relate the following functions to the graph details of ln x.

Here are the functions.

y = 3 ln x

y = -2 ln x

y = ln 2x

y = ln absolute value of x

y = ln (x-1)

y = 2 + ln x

Please help me with these problems as I am finding it real difficult to get the answer. Thanks for all your help and concern.

Regards
Salim Khan
.....................................
Hi, Salim,

Here is a general scheme for stuff like this:

Suppose you already have the picture for y = f(x).  Now you want to draw:

A.  y = f(x) + C.  Move the graph up C units. (down if C < 0)

B1.  y = f(x - a). Move the graph a units to the right.
B2.  y = f(x + a). Move the graph a units to the left.

C. y = cf(x).  Stretch the graph vertically by a factor of c.

D. y = - f(x).  Flip it upside down.

E. y = f(-x). Flip it left-right.

F. y = f(|x|)  [abs(x)].  Take the right side and copy it on the left as a mirror image.
......................
OK. Your examples.

y = 3 ln x -- use rule C.

y = -2 ln x -- use rules C, D.

y = ln 2x = ln2 + ln x.  Use rule A.

y = ln |x|.  Use rule F.  

y = ln (x-1) Use rule B1.

y = 2 + ln x -- Rule A again.