Answer:
[tex]\large a)\ f(x)=\dfrac{5}{9}(x-32)\\\\b)\ f^{-1}(x)=\dfrac{9}{5}x+32\\\\9^oC=48.2^oF[/tex]
Step-by-step explanation:
[tex]C=\dfrac{5}{9}(F-32)\\\\a)\ f(x)=\dfrac{5}{9}(x-32)\\\\b)\ C=\dfrac{5}{9}(F-32)\\\\\text{solve for F}\\\\\dfrac{5}{9}(F-32)=C\qquad\text{multiply both sides by 9}\\\\5(F-32)=9C\qquad\text{divide both sides by 5}\\\\F-32=\dfrac{9}{5}C\qquad\text{add 32 to both sides}\\\\F=\dfrac{9}{5}C+32\\\\f^{-1}(x)=\dfrac{9}{5}x+32[/tex]
[tex]c)\ \text{Put}\ x=9\ \text{to}\ f^{-1}(x):\\\\f^{-1}(9)=\dfrac{9}{5}(9)+32=\dfrac{81}{5}+32=16.2+32=48.2\\\\9^oC=48.2^oF[/tex]
