+0  
 
+1
133
1
avatar+545 

https://vle.mathswatch.co.uk/images/questions/question2241.png

 

 

winkwinkwink

lynx7  Mar 31, 2018
 #1
avatar+89781 
+2

f(x)  =  3x + p

g(x)  = px + 4

 

I think this is supposed to be f(g(x))

 

f(g(x))  =  3(px + 4) + p   =  3px + 12 + p

 

So

 

3px + (12 + p )    = 6x  + q

 

Equating coefficients, we have

 

3p   =  6  ⇒    p  =  2

  

12 +  p  =  q   ⇒  12 + 2  =  q   ⇒  14  =  q

 

If we interpret this as fg(x), we get that

 

(3x + p) (px + 4)  =  3px^2 + p^2x + 12x + 4p  =  3px^2 + (p^2 + 12)x + 4p

 

So

 

3px^2 + (p^2 + 12)x  + 4p  = 6x + q

 

Equating coefficients, we have

 

3p  =  0

p^2 + 12  =  6

4p  = q

 

 

But....the first equation  means p  =  0   and the second equation is impossible for any real p....so.....no solution  for fg(x)

 

 

cool cool cool

CPhill  Mar 31, 2018

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