Respuesta :

youssefjana646

Answer:

k^-1(x) = -1/128+16x

Domain = ]-infinity,-8[ U ]-8,infinity[

Range = ]-infinity,0[ U ]0,infinity[

Step-by-step explanation:

k(x) = -1/16x = -8

y = -1/16x = -8

x = -1/16y = -8

y = -1/128+16x

k^-1(x) = -1/128+16x

Answer:

D

Step-by-step explanation:

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