Question

The one-to-one function h is defined by
h(x)= 9x-8/3+8x
.
Find h^-1, the inverse of h. Then, give the domain and range of h^-1 using interval notation.

Answer 1

please and thank you =)

Answer 2

let y = 9x-8/3+8x
then 3y+8xy=9x-8
x(8y-9)=-3y-8
x= 3y+8/9-8y
h^-(x) = 3y+8/9-8y

Answer 3

i dont understand the domain and range

Answer 4

domain is all real numbers except 9/8
range all real numbers

Answer 5

would it be written with open or closed paranthesies

Answer 6

the answer asks for open or closed lol

Answer 7

open we cannot include infinity

Answer 8

im assuming the range is both open

Answer 9

for the domain are any closed so would it be???

(-infinity,8) (8,9) (9,infinity)

Answer 10

i think i got the range..just confused on the domain

Answer 11

domain is set for all function is defined and have finite value so R-{9/8}

Answer 12

the range is (-infinity,infinity)

Answer 13

yes

Answer 14

lol sorry just trying to understand this..i appreciate your help

Answer 15

the domain is the same as the range?

Answer 16

no in domain we cannot include the value 9/8 because for this value denominator becomes 0 and function not defined

Answer 17

so how i wrote it was right?

Answer 18

yes

Answer 19

thank you!!!!!!!