Question

Given that f(x)=2x/3 evaluate f^-1(6)

Answer 1

Answer:

9

Step-by-step explanation:

we know by definition that

$f^{-1}(f(x))=x\\\\f^{-1}(\frac{2x}{3})=x\\\\f^{-1}(x)=ax\\\\f^{-1}(\frac{2x}{3})=a(\frac{2x}{3})=x\\\\f^{-1}(x)=\frac{3x}{2}$

so now we evaluate

$f^{-1}(6)=\frac{3*6}{2}\\\\=3*3\\\\f^{-1}(6)=9$

and if we want to do an extra step

$f^{-1}(f(6))=6\\\\f^{-1}(\frac{2*6}{3})=f^{-1}(4)=\frac{3*4}{2}=6$

which works.