[tex]g(x)= \sqrt{x-5} , h(x)= x^{2} -6[/tex]
case (i):
the output of the first machine ([tex]\sqrt{x-5}[/tex]) is the input of the second machine, so we plug ([tex]\sqrt{x-5}[/tex]) or g(x) in function h and calculate, as follows:
[tex]h(g(x))=h(\sqrt{x-5})[/tex], now h is the function which squares whatever the input is and subtracts 6:
[tex]h(\sqrt{x-5})=(\sqrt{x-5}) ^{2}-6=x-5-6=x-11[/tex]
case (ii): we plug h in g:
[tex]g(h(x))= \sqrt{h(x)-5}= \sqrt{ x^{2} -6-5}= \sqrt{ x^{2} -11} [/tex]
a.
case1
[tex]h(g(x))=x-11=5
[/tex]
x=16
so for x=16, h(g(x))=5
case2
[tex]g(h(x))= \sqrt{ x^{2} -11}=5[/tex]
[tex] x^{2} -11=25[/tex]
[tex] x^{2} =36[/tex]
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
[tex]h(g(x))=x-11=-5[/tex]
x=11-5=6,
so for x=6, h(g(x))=-5
case2
[tex]g(h(x))= \sqrt{ x^{2} -11}=-5[/tex]
the square root cannot be a negative number
Answer:
A. order is g(h(x))
B. yes, for h(g(x)) only
