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# C3 watch

1. I have a few questions I'm stuck on and would appreciate some help.

We have f(x)= ln(3x - 6) when x is any real no. and x>2

How would i go about finding its range?

f-1=(ex + 6)/3

Since its range is f(x)'s domain, would the range be x>2? I have difficult in finding the range and domain sometimes. How would I find its domain?

y=ln(3x-6), to calculate where it cuts the x-axis I'd plug in y=0 and solve accordingly, but what if I wanted to calcualte it by means of translation, like for example, we know that lnx cuts the x-axis at 1, so ln3x would cut it 1/3, hence ln(3x-6) would cut the x-axis at 6 1/3. But that answer doesn't seem to correspond with the other, where have I went wrong?

And one more thing, how would I draw the graph of y=ln l3x - 6l.

Rep will be dished out in due course!

Thanks
2. (Original post by Dimez)
I have a few questions I'm stuck on and would appreciate some help.

We have f(x)= ln(3x - 6) when x is any real no. and x>2

How would i go about finding its range?

f-1=(ex + 6)/3

Since its range is f(x)'s domain, would the range be x>2? I have difficult in finding the range and domain sometimes. How would I find its domain?

y=ln(3x-6), to calculate where it cuts the x-axis I'd plug in y=0 and solve accordingly, but what if I wanted to calcualte it by means of translation, like for example, we know that lnx cuts the x-axis at 1, so ln3x would cut it 1/3, hence ln(3x-6) would cut the x-axis at 6 1/3. But that answer doesn't seem to correspond with the other, where have I went wrong?

And one more thing, how would I draw the graph of y=ln l3x - 6l.

Rep will be dished out in due course!

Thanks
domain of f is range of f^-1
range of f is domain of f^-1
if domain of f is x>2 then range of f^-1 is f^-1>2.
this can be seen by
f^-1=[e^x+6]/3=2+{e^x}/3
since e^x/3>0 for all x clearly f^-1 is >2 for all values.
the transformation ln(3x-6) depends on the order of the transformations.
As you note if we do the stretch then translation it does not correspond.
if we do translation then stretch we get
ln(x-6)cuts at x= 7
ln(3x-6) cuts at x=7/3
3. Thank you evariste.

Can you help me with the latter part please, drawing the graph. I understand that its got something to do with relecting on the y-axis. Do I draw y=ln(3x -6) for x>0 then reflect on the y -axis? If that is it, the graph would emerge somewhat weird.

Regards,

Dimez
4. (Original post by Dimez)
Thank you evariste.

Can you help me with the latter part please, drawing the graph. I understand that its got something to do with relecting on the y-axis. Do I draw y=ln(3x -6) for x>0 then reflect on the y -axis? If that is it, the graph would emerge somewhat weird.

Regards,

Dimez
The problem occurs at x<2, since 3x-6 is<0 and ln(3x-6) makes no sense.
The mod signs rid us of that problem.
x=2 will be assymptote.
when x=7/3 y=0
x=5/3 y=0
start at top left of graph paper. curve downwards so it goes through 5/3,0 and goes down further as you get closer to x=2
start at top right of graph paper, curve downwards so it goes through 7/3,0 and goes down further as you get closer to x=2.

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