Question

The table below shows some inputs and outputs of the invertible function f ff with domain all real numbers.

Answer 1

Answer: $f^{-1}(1)+f(-14)=20$

$f^{-1}(-2)=10$  

Step-by-step explanation:

The given table :

x: -14,-7,-12,9,10,-2

f(x):11,-12,5,1,-2,13

Since f is invertible ( given) , then $f^{-1}(x)$  exists.

Now , from table $f^{-1}(1)=9$  [ x= 9 corresponding to f(x) =1]

$f(-14)=11$                            [ f(x) = 11 corresponding to x=-14]

then, $f^{-1}(1)+f(-14)=9+11=20$

So, $f^{-1}(1)+f(-14)=20$

Also, x= 10 corresponding to f(x) =-2, then

$f^{-1}(-2)=10$