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If $P(x) = 4+2\sqrt{x+2}$ and $G(x) = 4-3x$, then what is the largest constant $a$ such that $P(G(a))$ is defined?

Guest Nov 15, 2014

Best Answer 

 #1
avatar+87301 
+13

P(G(a))  = 4 + 2√[(4-3x) +2] = 4 + 2√(6-3x )

Note that (6 - 3x) must be ≥ 0

So......

6 - 3x ≥ 0   add 3x to both sides

6 ≥ 3x        divide both sides by 3

2 ≥ x   →   x ≤ 2

So....."a" cannot be  > 2

 

CPhill  Nov 15, 2014
 #1
avatar+87301 
+13
Best Answer

P(G(a))  = 4 + 2√[(4-3x) +2] = 4 + 2√(6-3x )

Note that (6 - 3x) must be ≥ 0

So......

6 - 3x ≥ 0   add 3x to both sides

6 ≥ 3x        divide both sides by 3

2 ≥ x   →   x ≤ 2

So....."a" cannot be  > 2

 

CPhill  Nov 15, 2014
 #2
avatar
0

So, what do I write for a?

Guest Nov 15, 2014
 #3
avatar+87301 
+3

The largest value that "a" can assume is 2.....

 

CPhill  Nov 16, 2014

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