### Instruction: If $f(x)=\dfrac{a}{x+2}$, solve for the value of $a$ so that $f(0)=f^{-1}(3a)$.### Response: I know that $f(0)$ means I plug in $x=0$ into the function, so $f(0)=\dfrac{a}{0+2}=\dfrac{a}{2}$.
I also know that $f^{-1}(3a)$ means I plug in $3a$ into the inverse function, which is the function that undoes what $f$ does.
To find the inverse function, I can switch the roles of $x$ and $y$ in the original equation and solve for $y$.
So, if $y=\dfrac{a}{x+2}$, then $x=\dfrac{a}{y+2}$.
Solving for $y$, I get $y=\dfrac{a-2x}{x}$.
This is the inverse function, so $f^{-1}(x)=\dfrac{a-2x}{x}$.
Now, I can plug in $3a$ into this function and get $f^{-1}(3a)=\dfrac{a-2(3a)}{3a}$.
Simplifying, I get $f^{-1}(3a)=-\dfrac{5a}{3a}=-\dfrac{5}{3}$.
Now, I have $f(0)=f^{-1}(3a)$, so I can set them equal to each other and solve for $a$.
I get $\dfrac{a}{2}=-\dfrac{5}{3}$.
Multiplying both sides by $2$, I get $a=-\dfrac{10}{3}$.