graph of y=ln|3x-6|

can someone please explain to me why even thought the modulus is there, the graph can still have negative y values??

ln x is only valid on the reals for x>0, hence the modulus sign in this case ensures that all values of |3x-6| are >0. Look at the graph of ln x, it has negative y values. If the graph was a modulus of the whole thing, ie |ln(|3x-6|)| that would be a different matter!

Reply 2

john!!

erm. basically because ln of a number between 0 and 1 is negative, and the modulus doesn't do anything to numbers between 0 and 1.

Reply 3

AlphaNumeric

wacabac

hence the modulus sign in this case ensures that all values of |3x-6| are >0.

The mod makes

|3x-6| \geq 0

, because x=2 will still give ln 0.

Reply 4

futureaussiecto

thanks fella's

talking of modulae (plural i think lol),

find where these two lines intersect

y=4x+a and y=|x-a|+a

now in the mark scheme they reversed the signs inside the mod and did, 4x+a=(a-x)+a

can someone please explain why?? Ive never seen this before, is it in any of the textbooks??

Reply 5

topcrabcakes

y=4x+a and y=|x-a|+a

so we want:

4x + a = |x - a| + a

now if (x - a) > 0 then |x - a| = x-a

if (x - a) < 0 then |x - a| = -(x-a) = a - x

so then you solve:

4x + a = (x - a) + a = x (ie where (x - a) > 0)
x = 0, y = a

and:

4x + a = (a - x) + a (ie where (x - a) < 0)
x = a/5, y = 9a/5

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