Question

I need help with this question

Answer 1

`g(x)= √(x-5) , h(x)= x^(2) -6`

case (i):
the output of the first machine (
`√(x-5)`) is the input of the second machine, so we plug (
`√(x-5)`) or g(x) in function h and calculate, as follows:


`h(g(x))=h(√(x-5))`, now h is the function which squares whatever the input is and subtracts 6:


`h(√(x-5))=(√(x-5)) ^(2)-6=x-5-6=x-11`

case (ii): we plug h in g:


`g(h(x))= √(h(x)-5)= \sqrt{ x^(2) -6-5}= \sqrt{ x^(2) -11}`


a.

case1


`h(g(x))=x-11=5`

x=16

so for x=16, h(g(x))=5

case2


`g(h(x))= \sqrt{ x^(2) -11}=5`


`x^(2) -11=25`


`x^(2) =36`

x=-6 or x=6

for both these values, g(h(x))=5


so since her input was 6, the order was g(h(x)), that is h(x) was the input of g(x)


b.

case1


`h(g(x))=x-11=-5`

x=11-5=6,

so for x=6, h(g(x))=-5

case2


`g(h(x))= \sqrt{ x^(2) -11}=-5`

the square root cannot be a negative number



Answer:

A. order is g(h(x))

B. yes, for h(g(x)) only