+0

# Inverse?

0
187
2

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

Dec 4, 2018

#1
+1

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   Dec 4, 2018
#2
+12

Inverse?

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

$$\begin{array}{|rcll|} \hline f\left(f^{-1}(x) \right) &=& x \quad & | \quad x = 3a \\\\ f\left(f^{-1}(3a) \right) &=& 3a \quad & | \quad f^{-1}(3a) = f(0) \\\\ f\left(f(0) \right) &=& 3a \quad & | \quad f(0) = \dfrac{a}{0+2} = \dfrac{a}{2} \\\\ f\left(\dfrac{a}{2} \right) &=& 3a \quad & | \quad f\left(\dfrac{a}{2}\right) = \dfrac{a}{\dfrac{a}{2}+2} \\\\ \dfrac{a}{\dfrac{a}{2}+2} &=& 3a \\\\ \dfrac{1}{\dfrac{a}{2}+2} &=& 3 \\\\ \dfrac{a}{2}+2 &=& \dfrac{1}{3} \\\\ \dfrac{a}{2} &=& \dfrac{1}{3}-2 \\\\ \dfrac{a}{2} &=& -\dfrac{5}{3} \\\\ \mathbf{a} & \mathbf{=} & \mathbf{-\dfrac{10}{3}} \\\\ \hline \end{array}$$ Dec 4, 2018