October 21st, 2010, 09:23 AM  #1 
Newbie Joined: Oct 2008 Posts: 6 Thanks: 0  Making of equation
Hi there I'm having some problems with this one can't really figure it out couldn't attach the graphs so I wrote where they hit the y and x axis instead the first is a curve and the other is linear. Make the function y = x * f(g(x)) where the function y=f(g(x)) is defined by two graphs. One of the graphs y = f(u) (the graph is a curve that xaxis on 1 and the yaxis on just around 0.25 and 3.75) has the tangents where u = 1 and u = 3 and the other graph is named u = g(x) (the graph is linear and hits the yaxis on 2.5). The graph for the equations y = x* f(g(x)) has the tangent x = 1. Define the equation for the tangent. /Mathmen 
October 21st, 2010, 10:41 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
You should be able to upload an image file by using the "Upload attachment" button below the Submit button below the text box.

October 22nd, 2010, 04:39 AM  #3 
Newbie Joined: Oct 2008 Posts: 6 Thanks: 0  Re: Making of equation
Since it's a bmp file it won't work so I can't actually upload it =/ this is as far as I have come =/ y = f(u) u = g(x) if we incorporate u= g(x) in to y = f(u) we get y = f(g(x)) which is one step closer to y= x * f(g(x)) but from here I can't really use a valid formula that allows me to go any further. I was thinking y2  y1 = k(x2  x1) or also y = kx+b 
October 22nd, 2010, 05:52 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
There are various ways to convert your image file to a format that can be uploaded here or you could create a simple diagram that is roughly like the original. Alternatively, there are many sites that will let you upload your file (for free) so that you can link to it from here.

October 22nd, 2010, 07:29 AM  #5 
Newbie Joined: Oct 2008 Posts: 6 Thanks: 0  Re: Making of equation
Skipjack or if there is anyone else if it's possible could you check if my calculation are right? If they are then I finally managed to get the right answer after many hours of work :P After sitting with the question for a couple of hours and come to this answer, I would really appreciate if anyone can tell me if I made any type of mistake during the process because I'm quite confused if it should actually become y2 = x + f(g(1)) or y2 = x* f(g(1)) y2 –y 1 = k(x2 –x1) y2 – f(g(1)) = f’(g(1)) * g(1) (x2 1) y2 – f(g(1)) = f’(g(1)) * g(1) * x  f’(g(1)) * g(1) y2 = f’(g(1)) * g(1) * x  f’(g(1)) * g(1) + f(g(1)) y2 = x* f(g(1)) Thanks! 
October 22nd, 2010, 10:03 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
As your description is unclear, I'd like to see the diagram; is there a problem with uploading it to somewhere?


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