function notation and transformation of functions

bumblebee123

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can anyone help me to understand the answer to this question?

question: the function f is defined as f (x) = (x + 3) / (x - 1 )

find ff(x). give your answer in its simplest form.

any help would be really appreciated :)
 

ksdhart2

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One would naturally presume that \(ff(x)\) is function composition notation and means the same thing as \(f(f(x))\). That is to say, it's asking you to evaluate the function \(f(x)\) with \(f(x)\) passed as an argument. So just write out the given definition of \(f(x)\), replacing each \(x\) with \(f(x)\) and then in turn replace \(f(x)\) with its definition again:

\(\displaystyle f(f(x)) = \frac{f(x)+3}{f(x)-1} = \text{???}\)
 

pka

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can anyone help me to understand the answer to this question?
question: the function f is defined as f (x) = (x + 3) / (x - 1 ) find ff(x). give your answer in its simplest form.
I have a question. Was the question actually given as \(\displaystyle f\,f(x)~?\)
OR was it written as either \(\displaystyle f\circ f(x)\) or \(\displaystyle f\cdot f(x)~?\)
 

bumblebee123

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I have a question. Was the question actually given as \(\displaystyle f\,f(x)~?\)
OR was it written as either \(\displaystyle f\circ f(x)\) or \(\displaystyle f\cdot f(x)~?\)
it was given as ff( x ) :)
 

bumblebee123

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One would naturally presume that \(ff(x)\) is function composition notation and means the same thing as \(f(f(x))\). That is to say, it's asking you to evaluate the function \(f(x)\) with \(f(x)\) passed as an argument. So just write out the given definition of \(f(x)\), replacing each \(x\) with \(f(x)\) and then in turn replace \(f(x)\) with its definition again:

\(\displaystyle f(f(x)) = \frac{f(x)+3}{f(x)-1} = \text{???}\)
so f ( f (x) ) = ( ( ( x+3) / ( x -1 )) + 3 ) / ( ((x +3 ) / ( x - 1 ) )) - 1)
 

HallsofIvy

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Yes, now do the algebra to simplify that.
 

bumblebee123

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Yes, now do the algebra to simplify that.
( 4x / (x-1) ) / ( 4 / (x-1) ) = ( 4x^2 - 4x) / ( 4x - 4) = ( 4 ( x^2 - x ) ) / ( 4 ( x-1) ) = (x^2 - x ) / ( x -1 ) = ( x ( x - 1) ) / ( x -1 ) = x

so ff(x) = x which is correct

thanks :)
 

pka

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\(\displaystyle \begin{align*}\frac{{\frac{{x + 3}}{{x - 1}} + 3}}{{\frac{{x + 3}}{{x - 1}} - 1}} &= \frac{{\left( {\frac{{x + 3}}{{x - 1}} + 3} \right)(x - 1)}}{{\left( {\frac{{x + 3}}{{x - 1}} - 1} \right)(x - 1)}}\\& = \frac{{x + 3 + 3(x - 1)}}{{x + 3 - (x - 1)}}\\& = \frac{{4x}}{4}\\& = x \end{align*}\)
 
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