# draw a curve

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• Jan 26th 2011, 09:38 AM
paulaa
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 ?
• Jan 26th 2011, 10:47 AM
Archie Meade
Quote:

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 ?

$\displaystyle f(x)=ln2x$

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

$\displaystyle f(|1-x|)=ln[2(|1-x|)]$
• Jan 27th 2011, 03:19 PM
paulaa
then how will my curve look like ?
• Jan 27th 2011, 03:29 PM
pickslides
Quote:

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.
• Jan 27th 2011, 03:32 PM
paulaa
okej.

f(|1-x|)=ln[2(|1-x|)] is it the same as y=ln[2(|1-x|)]
?
• Jan 27th 2011, 04:13 PM
Archie Meade
Quote:

Originally Posted by paulaa
okej.

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

$\displaystyle y=ln(2|1-x|)\Rightarrow\ e^y=2|1-x|$

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

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

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

Also the graph is defined for positive and negative x as the modulus is being taken.
Therefore the graph is symmetrical about the asymptote.
• Jan 28th 2011, 02:52 AM
Krahl
If you're familiar with ln(x) you could also do it this way.
$\displaystyle f(x)=ln(x)$
Has a vertical asymptote at x=0 and f(x) = 0 at x = 1.
$\displaystyle f(x)=ln|x|$
makes the function even and symmetric about the y-axis.
$\displaystyle f(x)=ln|-x|$
Reflect the graph about the y-axis.
$\displaystyle f(x)=ln|-(x-1)|$
Shift the graph 1 unit to the right.
$\displaystyle f(x)=ln|-2(x-1)|=ln|-(x-1)|+ln2$
Shift the graph up ln2 units.