Inverse Functions

f(0)  =   a/2

To find the inverse, we have that

y =   a / [x + 2]

y  [ x + 2 ] = a

yx + 2y  =  a

yx =  a - 2y

x  =  [ a - 2y] / y       swap  x and y

y = [ a - 2x ] /x     =  the inverse = f^-1(x)

So

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

So  if f(0)  = f^-1(3a).....then

a/ 2   =  -5/3

a = -10/3

Inverse Functions

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

solve for the value of  \(a\) so that

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

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