Dec 14, 2009
423
1
0
Status
Pre-Dental
If F(x) = x and G(x) = square root (x-3) +2, what would the definition of H(x) have to be such that H(G(x)) = F(x) ?

answer is H(x) = (x-2)^2 +3

I read the solution manuals many times, but still don't understand. I need Ur help !
 
Jun 14, 2009
800
3
0
Status
Pre-Dental
If F(x) = x and G(x) = square root (x-3) +2, what would the definition of H(x) have to be such that H(G(x)) = F(x) ?

answer is H(x) = (x-2)^2 +3

I read the solution manuals many times, but still don't understand. I need Ur help !

So you need the function H(x) to be able to take in G(x) = (x-3)^1/2 +2 and put out x. Basically, H(x) has to undo whatever math is embedded in G(x).

Since you start with G(x) = (x-3)^1/2 + 2 and you want to work outside-in the first thing you want to do is subtract 2. Next you want to cancel out the square root by squaring it, so the new function is (x-2)^2. Now all that's left is x-3, so you add 3 to get your new function, (x-2)^2 + 3

G(x)---------------H(x)---------H(G(x))
(x-3)^1/2 +2------x------------(x-3)^1/2 +2
(subtract 2)-------x-2----------(x-3)^1/2
(square)----------(x-2)^2------(x-3)
(add 3)-----------(x-2)^2+3---x

voila.
H(x) = (x-2)^2 + 3, G(x) = (x-3)^1/2 + 2
H(G(x)) = ([(x-3)^1/2 +2]-2)^2 + 3
= [(x-3)^1/2]^2 + 3
= (x-3) + 3
= x = F(x)