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avatar+836 

If \(f(x)=\dfrac{a}{x+2}\), solve for the value of a so that \(f(0)=f^{-1}(3a)\).

 Dec 9, 2018
 #1
avatar+94499 
+1

Let's find the inverse of f(x)

 

y =   a / ( x + 2)

 

y (x + 2) = a

 

yx + 2y = a

 

yx = a - 2y      divide both sides by y

 

x =  [ a - 2y] / y    "swap" x and y

 

y = [ a - 2x] / x      and this is the inverse = f-1(x)

 

Note that f(0) =  a /2

 

And f-1 (3a) =   [ a - 2(3a) ] 3a  =  -5a / 3a =   -5/3

 

 

So   we want to solve this

 

a/2  = -5/3

 

a =  -10/3

 

 

cool cool cool

 Dec 9, 2018
 #2
avatar+20831 
+5

If

\(\large{f(x)=\dfrac{a}{x+2}}\),

solve for the value of a so that

\(\large{f(0)=f^{-1}(3a)}\).

 

\(\begin{array}{|rcll|} \hline f\left(~ f^{-1}(x) ~\right) &=& x \quad & | \quad x = 3a \\ f\left(~ f^{-1}(3a) ~\right) &=& 3a \quad & | \quad f^{-1}(3a) = f(0) \\ f\left(~ f(0) ~\right) &=& 3a \quad & | \quad f(0) = \dfrac{a}{0+2}=\dfrac{a}{2} \\ f\left(~ \dfrac{a}{2} ~\right) &=& 3a \quad & | \quad f\left(\dfrac{a}{2} \right) = \dfrac{a}{\dfrac{a}{2}+2}=\dfrac{2a}{a+4} \\ \dfrac{2a}{a+4} &=& 3a \\ \dfrac{2 }{a+4} &=& 3 \\ 2 &=& 3 (a+4) \\ 2 &=& 3a+12 \\ 3a &=& -10 \\ \mathbf{a} &\mathbf{=}& \mathbf{-\dfrac{10}{3}} \\ \hline \end{array}\)

 

laugh

 Dec 10, 2018

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