C3 HW help

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    Im stuck on this question:

    Show that the composite function gf is (8x^3 - 1) / (1 - 2x^3)

    f(x)= 1 - 2x^3
    g(x)=(3/x) - 4

    I subbed f into g but didnt get it
    any help appreciated


    also, solve gf(x)=0, if i set it = 0 then i cant bring anything across, do i have to factorise and cancel down the bottom maybe?
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    (Original post by not_lucas1)
    Im stuck on this question:

    Show that the composite function gf is (8x^3 - 1) / (1 - 2x^3)

    f(x)= 1 - 2x^3
    g(x)=(3/x) - 4

    I subbed f into g but didnt get it
    any help appreciated


    also, solve gf(x)=0, if i set it = 0 then i cant bring anything across, do i have to factorise and cancel down the bottom maybe?
    You need to put 1-2x^3 where the x is in g(x) and then put it all over the same denominator
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    (Original post by not_lucas1)
    Im stuck on this question:

    Show that the composite function gf is (8x^3 - 1) / (1 - 2x^3)

    also, solve gf(x)=0, if i set it = 0 then i cant bring anything across, do i have to factorise and cancel down the bottom maybe?
    0 times anything is 0 so when you multiply up by the denominator you just get the numerator =0
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    (Original post by tiny hobbit)
    You need to put 1-2x^3 where the x is in g(x) and then put it all over the same denominator
    i did and i didnt get the same answer
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    (Original post by not_lucas1)
    i did and i didnt get the same answer
    Did you get this? You'd only have to distribute and simplify.

    g\circ f(x)=g(f(x))=g(1-2x^{ 3 })=\frac { 3 }{ 1-2x^{ 3 } } -4=\frac { 3-4(1-2x^{ 3 }) }{ 1-2x^{ 3 } }
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    (Original post by MartyO)
    Did you get this? You'd only have to distribute and simplify.

    g\circ f(x)=g(f(x))=g(1-2x^{ 3 })=\frac { 3 }{ 1-2x^{ 3 } } -4=\frac { 3-4(1-2x^{ 3 }) }{ 1-2x^{ 3 } }
    No but I think you're correct. I didnt think to put 4/1 and then make it a common denominator (i think that is what you have done anyway)
 
 
 
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