Thread: 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 ?

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 ?

replace with

then how will my curve look like ?

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.

okej.

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

Originally Posted by paulaa

okej.

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

so there is a vertical asymptote at where the graph goes to

both to the left and right of

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

If you're familiar with ln(x) you could also do it this way.

Has a vertical asymptote at x=0 and f(x) = 0 at x = 1.

makes the function even and symmetric about the y-axis.

Reflect the graph about the y-axis.

Shift the graph 1 unit to the right.

Shift the graph up ln2 units.