SOLUTION: I have a couple that I am can not figure out thanks so much for your help g(x)= sqrt 5x-4 h(x)= 4x^2+7 Find: h(g(x)) g(x)=5x^2-4x _______ h(x)= sqrt x-7 find

Algebra ->  Rational-functions -> SOLUTION: I have a couple that I am can not figure out thanks so much for your help g(x)= sqrt 5x-4 h(x)= 4x^2+7 Find: h(g(x)) g(x)=5x^2-4x _______ h(x)= sqrt x-7 find      Log On


   

Question 547261: I have a couple that I am can not figure out
thanks so much for your help
g(x)= sqrt 5x-4
h(x)= 4x^2+7
Find: h(g(x))

g(x)=5x^2-4x
_______
h(x)= sqrt x-7
find the domain of : g(h(x))
Found 2 solutions by mathie123, stanbon:
Answer by mathie123(224) About Me  (Show Source):
You can put this solution on YOUR website!
when we are trying to find h(g(x)), what we are doing is really just substituting g(x) for any x we see in h(x). This means
h%28g%28x%29%29=4%28sqrt%285x-4%29%29%5E2%2B7
Since we need to do exponents first from BEDMAS, we know that the square root and the squared are opposites of each other, and therefore cancel each other out. This leaves:
h%28g%28x%29%29=4%285x-4%29%2B7
h%28g%28x%29%29=20x-16%2B7
h%28g%28x%29%29=20x-9


So for the second part...
g%28x%29=5x%5E2-4x
_______
h%28x%29=+sqrt%28x-7%29
We need to first find g(h(x)). Just as in the first question, we just substitute h(x) for any x we see in g(x).
This leaves:
g%28x%29=5%28sqrt%28x-7%29%29%5E2-4%28sqrt%28x-7%29%29
g%28x%29=5%28x-7%29-4%28sqrt%28x-7%29%29
Now the domain just means to see if there is any value of x that will make the above formula not work.
The first part of the formula (x-7) does not cause any problems.
The second part however does. We know that we cannot have a negative number under a square root (try it on your calculator... you will see it makes an error appear).
This means
x-7%3E=+0
x%3E=7
therefore the domain is any x such that x%3E=7

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
g(x)= sqrt 5x-4
h(x)= 4x^2+7
-------------------
Find: h(g(x)) = h[g(x)] = h[sqrt(5x-4)] = 4((5x-4)+7 = 20x-9
===========================
g(x)=5x^2-4x
h(x)= sqrt x-7
find the domain of : g(h(x))
g[h(x)] = g[sqrt(x-7)] = 5(x-4)-4(sqrt(x-7) = 5x-20-4sqrt(x-7)
=======================
Domain of g[h(x)] is All Real Numbers >= 7
============================================
Cheers,
Stan H.
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