1. ## draw a curve

i have got this information on a assigment:

f (x) = ln 2x , g(x) = cos(x - p / 2) , a (x) = 2x+ | x - 2 | ,

and the question is to draw the curve of y = f (|1- x |)

i think it will then be y = f (|1-ln2x|) right ?

2. Originally Posted by paulaa
i have got this information on a assigment:

f (x) = ln 2x , g(x) = cos(x - p / 2) , a (x) = 2x+ | x - 2 | ,

and the question is to draw the curve of y = f (|1- x |)

i think it will then be y = f (|1-ln2x|) right ?
$f(x)=ln2x$

$f(|1-x|)\Rightarrow$ replace $x$ with $|1-x|$

$f(|1-x|)=ln[2(|1-x|)]$

3. then how will my curve look like ?

4. Originally Posted by paulaa
then how will my curve look like ?
If you are unsure of the general shapes of the abs value and log functions, you can make a table of values and solve for any intercepts/asymptotes.

5. okej.

f(|1-x|)=ln[2(|1-x|)] is it the same as y=ln[2(|1-x|)]
?

6. Originally Posted by paulaa
okej.

f(|1-x|)=ln[2(|1-x|)] is it the same as y=ln[2(|1-x|)]
?
$y=ln(2|1-x|)\Rightarrow\ e^y=2|1-x|$

$e^{-\infty}\rightarrow\ 0$

so there is a vertical asymptote at $x=1$ where the graph goes to $-\infty$

both to the left and right of $x=1.$

Also the graph is defined for positive and negative x as the modulus is being taken.
Therefore the graph is symmetrical about the asymptote.

7. If you're familiar with ln(x) you could also do it this way.
$f(x)=ln(x)$
Has a vertical asymptote at x=0 and f(x) = 0 at x = 1.
$f(x)=ln|x|$
makes the function even and symmetric about the y-axis.
$f(x)=ln|-x|$
Reflect the graph about the y-axis.
$f(x)=ln|-(x-1)|$
Shift the graph 1 unit to the right.
$f(x)=ln|-2(x-1)|=ln|-(x-1)|+ln2$
Shift the graph up ln2 units.