SOLUTION: More problem with functions! Some I get easily and some....... s(x)=2^x h(x)=x^2 find s(h(x)) My solution was 2x^2 but apparently is wrong one!

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Question 214821: More problem with functions! Some I get easily and some.......
s(x)=2^x h(x)=x^2 find s(h(x))
My solution was 2x^2 but apparently is wrong one!
Found 2 solutions by drj, RAY100:
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
s%28x%29=2%5Ex+ h%28x%29=x%5E2 Find s(h(x))

Substitute h(x) into s(x) to yield the following:

s%28h%28x%29%29=2%5E%28h%28x%29%29=2%5E%28x%5E2%29

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

Answer by RAY100(1637) About Me  (Show Source):
You can put this solution on YOUR website!
s{x} = 2^(x)
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s{h(x)} =2^{h(x)} = 2^{x^2} =2^(x^2)
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Just do direct substitution,,,(it's a shame we use so many x"s)
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.for example,,,y=f(z) = {z+1}
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f(4) = {(4) +1},,,,just let all z's = 4
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then you try to simplify, in this case f(4) = 5,,,makes sense
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