If \(f(x)=\dfrac{a}{x+2}\), solve for the value of \(a\) so that \(f(0)=f^{-1}(3a)\).
f(0) = a/2
Find the inverse.....get x by itself
y = a / (x + 2)
yx + 2y = a
yx = a - 2y divide both sides by y
x = [ a - 2y] / y "swap" x and y
y = [ a - 2x] / x
f-1(x) = [ a - 2x ] / x this is the inverse
f-1 (3a) = [ a - 2(3a) ]/ (3a) = - 5a / 3a = -5/3
And this equals f(0).....so
a/2 = -5/3 multiply both sides by 2
a = -10/3
Inverse?
If \(\large{f(x)=\dfrac{a}{x+2}}\), solve for the value of \(\large{a}\) so that \(\large{f(0)=f^{-1}(3a)}\).
\(\begin{array}{|rcll|} \hline f\left(f^{-1}(x) \right) &=& x \quad & | \quad x = 3a \\\\ f\left(f^{-1}(3a) \right) &=& 3a \quad & | \quad f^{-1}(3a) = f(0) \\\\ f\left(f(0) \right) &=& 3a \quad & | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\\\ f\left(\dfrac{a}{2} \right) &=& 3a \quad & | \quad f\left(\dfrac{a}{2}\right) = \dfrac{a}{\dfrac{a}{2}+2} \\\\ \dfrac{a}{\dfrac{a}{2}+2} &=& 3a \\\\ \dfrac{1}{\dfrac{a}{2}+2} &=& 3 \\\\ \dfrac{a}{2}+2 &=& \dfrac{1}{3} \\\\ \dfrac{a}{2} &=& \dfrac{1}{3}-2 \\\\ \dfrac{a}{2} &=& -\dfrac{5}{3} \\\\ \mathbf{a} & \mathbf{=} & \mathbf{-\dfrac{10}{3}} \\\\ \hline \end{array}\)