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Math Help - Composite Functions

  1. #1
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    Composite Functions

    Will someone please solve the following composite function for g(x):

    g[f(x)] = x-1; given f(x) = sqrt(x+2)

    I believe the answer is (x^2) - 3, but I cannot remember how to solve using basic algebra.

    TIA,
    J. Lane
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Knowledge View Post
    Will someone please solve the following composite function for g(x):

    g[f(x)] = x-1; given f(x) = sqrt(x+2)

    I believe the answer is (x^2) - 3, but I cannot remember how to solve using basic algebra.

    TIA,
    J. Lane
    your answer is correct. loosely speaking there is no real algebraic method to do this, it is more like a pattern recognition problem. you should just realize that to go from \sqrt{x + 2} to x - 1 you would square and subtract 3
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  3. #3
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    Ok then, for the following composite function f[g(x)] = [(x^4)+(x^2)]/[1+(x^2)]; given g(x) = 1 - (x^2):

    How would you solve for f(x) using "pattern recognition"? I am certain there is an appropriate method to solve using algebra because I remember learning it back in the 8th grade, but as the saying goes, "if you don't use it, you lose it."

    Thank you again,
    J. Lane
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Knowledge View Post
    Ok then, for the following composite function f[g(x)] = [(x^4)+(x^2)]/[1+(x^2)]; given g(x) = 1 - (x^2):

    How would you solve for f(x) using "pattern recognition"? I am certain there is an appropriate method to solve using algebra because I remember learning it back in the 8th grade, but as the saying goes, "if you don't use it, you lose it."

    Thank you again,
    J. Lane
    hmm, i believe you are right. i have forgotten the method also. i will try to look it up and get back to you. the "pattern recognition" i talk about is probably using this method without realizing it. i do that sometimes. because i do something over and over, i tend to internalize it and end up doing it "naturally"--forgetting the original method or formula in the process.

    here i get f(x) = \frac {(1 - x)(2 - x)}{2 - x}

    again, recognizing patterns:

    note that we have f(g(x)) = \frac {x^2(1 + x^2)}{1 + x^2}

    for 1 - x^2 \to x^2, take the negative and add 1, hence you get 1 - x

    for 1 - x^2 \to 1 + x^2, take the negative and add 2, hence 2 - x
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