waloshin macrumors 68040 Original poster 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 Question number 3.doc 27 KB · Views: 129
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 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 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.
D darklyt macrumors regular 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.)