Results 1 to 3 of 3

Math Help - Algebra 2 Composition of Functions?

  1. #1
    Newbie
    Joined
    Jul 2012
    From
    USA
    Posts
    7

    Algebra 2 Composition of Functions?

    If g(f(x))= 2x+1, f(x)= (1/4)x-1, then what does g(x) equal? The answer is supposed to be 8x+9. Can anyone explain to me clearly step by step how to do this? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2012
    From
    AZ
    Posts
    616
    Thanks
    97

    Re: Algebra 2 Composition of Functions?

    "Fit" the expression of f(x) into g(f(x)):

    8f(x) = 8(\frac{1}{4}x - 1) = 2x - 8

    8f(x) + 9 = (2x - 8) + 9 = 2x + 1 = g(f(x))

    Therefore, g(x) = 8x + 9.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,659
    Thanks
    600

    Re: Algebra 2 Composition of Functions?

    Hello, woahitzme!

    \text{If }g(f(x))\:=\: 2x+1,\;\; f(x)\:=\:\tfrac{1}{4}x-1
    . . \text{then what does }g(x)\text{ equal?}
    \text{The answer is supposed to be: }\,8x+9.

    We are told that: . f(x) \:=\:\tfrac{1}{4}x-1

    Then: . g\left(\tfrac{1}{4}x-1\right) \:=\:2x+1

    Assume that g(x) is a linear function: . g(x) \:=\:ax+b

    Then: . g\left(\tfrac{1}{4}x-1\right) \:=\:a\left(\tfrac{1}{4}x-1\right) + b \;=\;2x + 1

    We have: . \tfrac{a}{4}x - a + b \:=\:2x+1 \quad\Rightarrow\quad \tfrac{a}{4}x + (b-a) \:=\:2x+1

    Equate coefficients: . \begin{array}{ccc}\dfrac{a}{4} &=& 2 \\ b-a &=& 1 \end{array}

    Hence: . a \,=\,8,\;b\,=\,9

    Therefore: . g(x) \:=\:8x + 9
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Composition of Two Functions
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 14th 2011, 11:26 AM
  2. composition of functions
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: May 27th 2010, 12:10 PM
  3. Composition & Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: November 13th 2008, 12:44 PM
  4. Composition of Two Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 9th 2008, 06:44 PM
  5. Composition of Two Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 9th 2008, 03:53 PM

Search Tags


/mathhelpforum @mathhelpforum