Math - Algebra II

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Hello, i'm currently doing functions and I was fine until I got to multiplication. I really don't know what i'm doing, and would really appreciate a step by step on a few questions.

f(x) = -3x + 9

g(x) = 8x + 7

Find g(f(x))

I got -24x + 16.

Another :

f(x) = -3x + 9

g(x) = 8x + 7

Find f(g(x))

Thanks in advance.

  • Math - Algebra II -

    no, one term was right, but your constant is wrong.

    do it this way :

    f(x) = -3x + 9
    g(x) = 8x + 7

    then g(f(x))
    = g(-3x+9)
    = 8(-3x+9) + 7
    = -24x + 72 + 7
    = -24x + 79

    and f(g(x))
    = f(8x+7)
    = -3(8x+7) + 9
    = -24x - 21 + 9
    = -24x - 12

  • Math - Algebra II -

    Thank you so much! I've been sitting here for hours. Only 4 to go. Thanks :)

  • Math - Algebra II -

    2 down.

    I'm having trouble here

    f(x) = 4x + 2

    h(x) = -7x - 5

    Find f(h(x))

    I tried doing it how it was done above, but i'm not getting an answer listed.

    = f(-7x - 5)
    = 4x(-7x-5)+ 2
    = 28x^2-20+2
    = 28x^2-18

    Where am I going wrong :| ?

    ALSO -

    f(x) = 4x-3

    g(x) = x2-2

    Find f(3a-4)

  • Math - Algebra II -

    from
    = f(-7x - 5) to
    = 4x(-7x-5)+ 2
    you are replacing the x, so don't leave the x there.
    should be
    = 4(-7x-5)+ 2
    = -28x - 20 + 2
    = -28x - 18


    for
    f(x) = 4x-3

    g(x) = x2-2

    Find f(3a-4)
    I will assume g(x) = x^2 - 2

    then f(3a-4)
    = 4(3a-4) - 3
    = 12a - 16 - 3
    = 12a - 19

    the x in f(x) was merely replaced by 3a-4, so you have to replace it everywhere that you see an x.

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