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You have linear functions p(x) and q(x). You know p(2)=3, and p(q(x))=4x+7 for all x. Find q(-1).

 

I don't know where to start for this problem, please help!!

 

XD

 Apr 7, 2020

Best Answer 

 #2
avatar+24983 
+2

You have linear functions p(x) and q(x).
You know p(2)=3, and p(q(x))=4x+7 for all x.
Find q(-1).

 

My attempt:

 

linear functions

\(\text{Let $p(x)=ax+b$} \\ \text{Let $q(x)=ux+v$}\)

 

\(\begin{array}{|rcll|} \hline && \boxed{p(x)=ax+b \\ p(2) = a*2 + b \quad | \quad p(2) = 3 \\ 3 = 2a + b \\ \mathbf{b = 3-2a} } \\ p(x) &=& ax+b \quad &| \quad x=q(x) \\ p\Big(q(x)\Big) &=& aq(x)+b \quad &| \quad q(x)=ux+v \\ p\Big(q(x)\Big) &=& a(ux+v)+b \\ p\Big(q(x)\Big) &=& aux+av+b \quad &| \quad p\Big(q(x)\Big) = 4x+7 \\ 4x+7 &=& (au)x+(av+b) \quad &| \quad \text{compare} \\ && \boxed{\mathbf{au = 4},\\ ~\\ \mathbf{av+b = 7} \quad | \quad b = 3-2a \\ av+3-2a = 7 \\ av-2a = 4\\ a(v-2) = 4 \quad | \quad 4 = au \\ a(v-2)=au \\ \mathbf{v-2 = u} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline q(x) &=& ux+v \quad &| \quad u = v-2 \\ q(x) &=& (v-2)x+v \quad &| \quad x=-1 \\ q(-1) &=& (v-2)(-1)+v \\ q(-1) &=& -v+2 +v \\ \mathbf{q(-1)} &=& \mathbf{2} \\ \hline \end{array} \)

 

laugh

 Apr 7, 2020
 #1
avatar
0

q(-1) = 17.

 Apr 7, 2020
 #2
avatar+24983 
+2
Best Answer

You have linear functions p(x) and q(x).
You know p(2)=3, and p(q(x))=4x+7 for all x.
Find q(-1).

 

My attempt:

 

linear functions

\(\text{Let $p(x)=ax+b$} \\ \text{Let $q(x)=ux+v$}\)

 

\(\begin{array}{|rcll|} \hline && \boxed{p(x)=ax+b \\ p(2) = a*2 + b \quad | \quad p(2) = 3 \\ 3 = 2a + b \\ \mathbf{b = 3-2a} } \\ p(x) &=& ax+b \quad &| \quad x=q(x) \\ p\Big(q(x)\Big) &=& aq(x)+b \quad &| \quad q(x)=ux+v \\ p\Big(q(x)\Big) &=& a(ux+v)+b \\ p\Big(q(x)\Big) &=& aux+av+b \quad &| \quad p\Big(q(x)\Big) = 4x+7 \\ 4x+7 &=& (au)x+(av+b) \quad &| \quad \text{compare} \\ && \boxed{\mathbf{au = 4},\\ ~\\ \mathbf{av+b = 7} \quad | \quad b = 3-2a \\ av+3-2a = 7 \\ av-2a = 4\\ a(v-2) = 4 \quad | \quad 4 = au \\ a(v-2)=au \\ \mathbf{v-2 = u} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline q(x) &=& ux+v \quad &| \quad u = v-2 \\ q(x) &=& (v-2)x+v \quad &| \quad x=-1 \\ q(-1) &=& (v-2)(-1)+v \\ q(-1) &=& -v+2 +v \\ \mathbf{q(-1)} &=& \mathbf{2} \\ \hline \end{array} \)

 

laugh

heureka Apr 7, 2020
 #3
avatar+111329 
+1

Very nice, heureka  !!!!

 

 

cool cool cool

CPhill  Apr 7, 2020
 #4
avatar+24983 
+2

Thank you, CPhill !

 

laugh

heureka  Apr 7, 2020

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