waloshin macrumors 68040 Original poster Oct 9, 2008 3,562 394 Apr 6, 2009 #1 On question number 3 , is that a typo because I dont understand the question. If not could you give me an example on how to do it? Attachments Question number 3.doc 27 KB · Views: 120
On question number 3 , is that a typo because I dont understand the question. If not could you give me an example on how to do it?
waloshin macrumors 68040 Original poster Oct 9, 2008 3,562 394 Apr 6, 2009 #3 I understand the basics: If f(x) is = 6x+1 and g(x) is = x+9 Then find f(x) + g(x) 6x + 1 + x + 9 = 7x + 10
I understand the basics: If f(x) is = 6x+1 and g(x) is = x+9 Then find f(x) + g(x) 6x + 1 + x + 9 = 7x + 10
Shaun.P macrumors 68000 Jul 14, 2003 1,601 24 Omicron Persei 8 Apr 6, 2009 #4 For f(x) = g(x) let the two functions equal each other. That means 2*x^2 - x [f(x)] = 3*x + 6 [g(x)]. You need to move everyone to one side, such that it will equal 0. Simplify and factorise. E.g: 2*x^2 - x = 3*x + 6 Move everything to the LHS: => 2*x^2 - 4*x - 6 = 0 Divide by common factor of 2: => x^2 -2*x - 3 =0 This factorises to: => (x-3)(x+1)=0 [Hopefully you will know how to factorise this]. => x=3 or x=-1.
For f(x) = g(x) let the two functions equal each other. That means 2*x^2 - x [f(x)] = 3*x + 6 [g(x)]. You need to move everyone to one side, such that it will equal 0. Simplify and factorise. E.g: 2*x^2 - x = 3*x + 6 Move everything to the LHS: => 2*x^2 - 4*x - 6 = 0 Divide by common factor of 2: => x^2 -2*x - 3 =0 This factorises to: => (x-3)(x+1)=0 [Hopefully you will know how to factorise this]. => x=3 or x=-1.
jodelli macrumors 65816 Jan 6, 2008 1,219 4 Windsor, ON, Canada Apr 6, 2009 #5 x=3 unless I'm reading it wrong edit: ShaunPriest got it.
D darklyt macrumors regular Jun 5, 2007 163 43 Apr 6, 2009 #6 Based on ShaunPriest's answer, you're just solving for the points of intersection. If you graph the functions together, the solutions are where the graphs of the functions cross one another. (Just so you have another interpretation.)
Based on ShaunPriest's answer, you're just solving for the points of intersection. If you graph the functions together, the solutions are where the graphs of the functions cross one another. (Just so you have another interpretation.)
Shaun.P macrumors 68000 Jul 14, 2003 1,601 24 Omicron Persei 8 Apr 7, 2009 #7 waloshin do you understand now?