SOLUTION: I seem to have a problem with functions! n(x)=3^x m(t)=SQRT(t)+1 find m(n(x)) My solution was SQRT(3^x)+1. Please help!

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Question 214823: I seem to have a problem with functions!
n(x)=3^x m(t)=SQRT(t)+1 find m(n(x))

My solution was SQRT(3^x)+1.
Please help!
Found 2 solutions by Earlsdon, Theo:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
I see nothing wrong with your solution!
n%28x%29+=+3%5Ex
m%28t%29+=+sqrt%28t%29%2B1
Find:
m%28n%28x%29%29
m%28n%28x%29%29+=+m%283%5Ex%29
m%283%5Ex%29+=+sqrt%283%5Ex%29%2B1
highlight%28n%28m%28x%29%29+=+sqrt%283%5Ex%29%2B1%29

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
n(x) = 3^x
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m(t) = sqrt(t) + 1
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find m(n(x))
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substitute n(x) for t in the equation of m(t).
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since n(x) = 3^x, this means that:
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m(t) = sqrt(t) + 1 becomes:
m(3^x) = sqrt(3^x) + 1
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that should be your equation.
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to prove this is true, substitute any value for x and solve.
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let x = 8 (chosen at random)
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n(x) = 3^x = 3^8 = 6561
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m(3^x) = sqrt(3^x) + 1 = sqrt(3^8)) + 1 = sqrt(6561) + 1 = 81 + 1 = 82
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m(t) = sqrt(t) + 1
let t = 6561
you get:
m(t) = sqrt(t) + 1 = sqrt(6561) + 1 = 81 + 1 = 82
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you get the same answer whether you solve for:
m(n(x)) = sqrt(n(x)) + 1 directly, or whether you solve for:
n(x) first and then use that value in m(t).
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the equations are equivalent.
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the key to solving these types of problems is to substitute the function for the variable and then solve.
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check this website out.
It has some examples that may help you to understand what is going on.
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http://www.regentsprep.org/Regents/math/algtrig/ATP5/EvaluatingFunctions.htm
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just copy that line into your browser address bar and hit the return.
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it's short and sweet so take the time to look at it if you can.
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fyi,
your solution was correct.
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