Graphing question (1 Viewer)

[Aysce]

Q. Sketch y= ln [f(x)]

Not sure how to do this. Any help + a brief explanation will be appreciated =)

 

[Carrotsticks]

Strange question, they essentially gave us that the equation of the curve is:



So drawing the ln curve would turn into sketching:



And then just add/subtract the curves.

 

[My Name is Khan]

I did a similar question a few days ago in one of the past papers...I think it was Sydney Grammer Trial 2011

 

[Aysce]

Strange question, they essentially gave us that the equation of the curve is:



So drawing the ln curve would turn into sketching:



And then just add/subtract the curves.

Is there a way to do it without finding the equation of the curve? I watched your video but I wasn't completely sure.

 

[My Name is Khan]

Is there a way to do it without finding the equation of the curve? I watched your video but I wasn't completely sure.

which video?

 

[Carrotsticks]

Left branch:

For simplicity's sake, I will call the new curve g(x).

Can't ln any values less than or equal to 0, so we ignore the part of the left branch that goes under the y axis.

We know that for the curve y=ln(x), when 0 < x < 1, - infinity < y < 0 (0 from the negative side btw), so likewise the left branch will behave similarly.

As y -> 1 for f(x), y -> 0 (negative side) for g(x).

As y -> 0 for f(x), y -> - infinity for g(x)

Right branch:

We know ln(infinitely large number) = infinitely large number still (but significantly smaller), so the asymptote at x=1 is preserved. The curve f(x) approaches 1 for large x, so g(x) will approach 0 for large x, since ln(1) = 0.

So our final answer is going to still look like a hyperbola, but with asymptotes at x=1, x=-1 and y=0.

 

[seanieg89]

Is there a way to do it without finding the equation of the curve? I watched your video but I wasn't completely sure.

You can't find "the equation" of that curve, many different equations will have identical looking sketches.

 

[My Name is Khan]

You can't find "the equation" of that curve, many different equations will have identical looking sketches.

+1

 

[Carrotsticks]

It's not the exact equation because we have infinitely many curves with those asymptotes and having those intercepts, but Aysce to help you when you're in trouble you can 'guess' what the curve looks like, then work with your approximation.

In reference to my post, the curve 'looks' like y=(x+1)/(x-1) so you can work with that curve as a bit of a guide.

Do NOT automatically assume that it's the actual equation because the actual equation could be anything for all we know. So do NOT make the mistake of using values from y=(x+1)/(x-1) and then using it in your final answer. I've seen people do this too many times. The given curve looks like y=e^x, so they automatically label the curve y=e^x and work with that equation instead. It's a GUIDE, not the actual thing.

For example if a curve 'looks' like the line y=x at some point and I have to draw ln(f(x)), then I 'compare' it to the curve y=ln(x) appropriately.

 

[D94]

Very easy method.

You let y = ln(u) and u = "that curve" (as a function of x). In the end, we match up the y and x points to form the new graph.

We need to test important points on the u-x curve (ie. the given curve), so in the above example, x = "very large positive number", x = 1, x = 0, x = -1, and x = "very large negative number". Then we find its corresponding u value, then we sub that u value into the y = ln(u) equation to get the corresponding y value.

Now we have the x and y points, and we can draw the curve.

 

[Aysce]

Ah okay I get it now. Thank you all who contributed =)