help

$\text{Well the brute force way is }\\ y=\dfrac{a}{x+2}\\ \dfrac 1 y = \dfrac{x+2}{a}\\ x = \dfrac a y - 2\\ f^{-1}(x) = \dfrac{a}{x} - 2$

$f(0) = f^{-1}(3a)\\ \dfrac{a}{2} = \dfrac{a}{3a}-2\\ \dfrac a 2 = \dfrac 1 3 - 2 = -\dfrac 5 3\\ a = - \dfrac{10}{3}$

If  $f(0)=f^{-1}(3a)$  ,  solve for the value of a so that    $f(0)=f^{-1}(3a)$

This looks kinda interesting   -  I will assume that a is a constant.

$let\;\\ y=\dfrac{a}{x+2} \quad x \ne-2\\ x+2=\dfrac{a}{y}\\ x=\dfrac{a}{y}-2\\ f^{-1}(x)=\dfrac{a}{x}-2\\~\\ f^{-1}(3a)=\dfrac{a}{3a}-2=\frac{1}{3}-2=\frac{-5}{3}\\ $

$f(0)=\frac{a}{2}$

if

$f(0)=f^{-1}(3a)\\ \frac{a}{2}=\frac{-5}{3}\\ a=\frac{-5*2}{3}\\ a=\frac{-10}{3}\\ a=-3\frac{1}{3}$

If

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

solve for the value of $\mathbf{a}$ so that

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

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