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Hi guys, as part of a question I had to sketch y=[ln(x)]/x which I easily completed, The next part of the question said to sketch y=x/ln(x), without any further calculation, i.e. no calculus. How do I go about sketching the second function when I already have sketched my first function? Also...

omg, thanks man!

Hi guys, this is the question I had to do: Q. Draw careful sketches of the following, clearly showing asymptotes. DO NOT USE CALCULUS. (g). y= (2x^2 + 4x + 3)/(x^2 - 1) So for this question, there's an asymptote at y=2, and x=1 and x=-1, no x-intercepts, and passes through (0,-3). The guide...

For one step of a certain question, I have to find: for what value of k, does y=kx become tangent to y=e^(-x) How do I go about doing this? Thanks guys

thanks, but just wondering, are there any other methods to get the asymptotes?

Hi guys, here's the question I'm having trouble with: 5x^2 - y^2 + 4xy = 18 ---(1), I have to sketch it. Here's what I did: --------------------------------------- 10x - 2yy' + 4y + 4xy' = 0 5x - yy' + 2y + 2xy' = 0 y' = [(2y+5x)/(y-2x)] As 2x-y--> 0, y'-->inf therefore, y=2x, sub in ---(1)...

cool thanks, I looked at the graph of a derivative that had a point make it equal 0/0, and it was really weird, the graph just jumped up to a random point on the same vertical line..

lol, that's crazy talk, youre a crap load smarter than me... mebbe i just can't explain very well

Ok, textbook says, if y' = [f'(x)/($$$)], and $$$= 0 , but f'(x) doesn't equal zero, then y' --> infinity, i.e. vertical tangent. Looks like in the, y=inv.sin(f(x), f'(x) happened to equal zero... I guess in an exam you would just guess that f'(x) equals zero, because you can't have a vertical...

Quick question: What is the difference between UNDEFINED (1 divided by 0) and INDETERMINABLE ( 0/0 or infinity/infinity, or 0*infinity, or infinity MINUS infinity)??? Thanks

Hi guys just need a bit of help with this one, here's how it goes: f(x) is an odd function, as x--> +/- inf, f(x) --> 0, there is a MAXIMUM at (1,1) and a MINIMUM at (-1,-1) and a point of inflexion at (0,0). I had to sketch g(x) = inv.sin(f(x)). Here's what I did...

Thankyou

So umm... how would that apply to y = inv.sin (cos x)?? And so if the domain is ALL REAL X, you would just use inv.sine(1) and inv.sine(-1) to find the domain??

Yeah is the range, but how one earth can you get the range by just doing: inv.sin(0) = 0 & inv.sin(1) = pi/2. Like wtf, why is it that you can figure out the range of the function through the inverse sine of "0" and "1". Also another example of this wierd method: y= inv.sin (x^2 -1) Quoted...

Basically I want to know is how did the textbook calculate the range using inv.sin(0) and inv.sin (1)

regarding the range of the function, all I did was sub various points in, e.g. x=0, x=1, x=10^20. However the textbook had a much faster method, but it wasn't all that clear to understand. It went: "range is 0<y<= (pi/2), due to inv.sin(0)=0 and inv.sin(1)=pi/2 I don't really understand where...

... the problem is... I am doing 4u... :S missed out on too much class work so I'm catching up at home... but probably should drop it anyways Thanks anyway And also, for x^2 = -2, how can you take the root of a negative? Or do you just solve the graph through graphing rather than algebraically?

-x^2 - 1 <= 1 x^2 <= -2 Uh oh... But when I graph -x^2 -1, it is <= 1 for ALL x, but then with x^2, it is not at all <=-2, so wtf is going on?? What the heck do I with the left end, just ignore it? x^2 + 1 > 1 x^2 > 0 help please

Hi, I have to find the domain and range of g(x)= inv. sin [1/(x^2 + 1)] Here's what i did: domain of y=inv.sin(x) is -1<(x)<1 therefore domain of g(x): -1 < [1/(x^2 +1)] < 1 -(x^2 +1) < 1 < (x^2 + 1) So... what's the correct way of doing it, other than randomly subbing in points? Thanks guys

So all you do is take the derivate of the exponent than multiply it by the original??