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Two functions f(x) and g(x) need a walk through if possible Watch

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    I have a question on functions but its a one off so wondering if I could get any help.
    Ive attached a word doc as I typed it on there as it has subscript and power button.

    M question starts as: if two functions and are defined as follows, () f x() gx
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  1. File Type: docxIf two functions are defined as follows.docx (12.9 KB, 24 views)
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    if it's fg(x) you put g into f and gf(x) is f into g.
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    Which part you stuck on? For a) you just plug in 0.1
    For b) you'd first find g(f(x)) and then plug in 0.1 or easier you have your answer to f(0.1) just plug that into g.
    c) find the inverse, by rearranging to find x, Same with d.
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    Thanks but I don't understand what you mean. The question was given to me but I need to do it before Monday in the format shown
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    I couldn't see what f(x) should be in the document but it goes as follows:
    say f(x) = 2x+5 and g(x) = x^2 and you are asked to solve f(g(5))
    then first solve g(5) = 5^2 = 25
    then plug the output of g into f(x) as the x parameter
    so f(25) = 2(25)+5 = 55 = f(g(5))
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    (Original post by Bulldog1965)
    Thanks but I don't understand what you mean. The question was given to me but I need to do it before Monday in the format shown
    For f(0.1) it wants you to substitute in the value  x=0.1 into the equation

    For part b, you may also seen this written as gf(0.1), it wants you to substitute in the value of f(0.1) for the x value into the equation g(x). As you already know the value of f(0.1) from the first question, you can just substitute that value straight in.

    For parts c and d, it wants you to find the inverse of those functions. The easier way probably is to rearrange the equation for x, then switch the signs around at the end.

    Like this for example,
     f(x)=\dfrac{7} {3x-8}

 y=\dfrac{7} {3x-8}

y(3x-8)=7

3x-8=\dfrac{7} {y}

3x=\dfrac{7} {y}-8

x=\dfrac{7} {3y}-{\frac{8} {3}} 

f^{-1}(x)=\dfrac{7} {3x}-{\dfrac{8} {3}}

    Does that make sense?
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    Great info guys I sorted it and it all fell into place thanks very much
 
 
 
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