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 ofy=f(|1-x|)

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

Printable View

- Jan 26th 2011, 09:38 AMpaulaadraw a curve
i have got this information on a assigment:

*f*(*x*) = ln 2*x*,*g*(*x*) = cos(*x*- p / 2) , a (*x*) = 2*x*+ |*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 AMArchie Meade
- Jan 27th 2011, 03:19 PMpaulaa
then how will my curve look like ?

- Jan 27th 2011, 03:29 PMpickslides
- Jan 27th 2011, 03:32 PMpaulaa
okej.

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

? - Jan 27th 2011, 04:13 PMArchie Meade
$\displaystyle y=ln(2|1-x|)\Rightarrow\ e^y=2|1-x|$

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

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 AMKrahl
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.