Answer:
Step-by-step explanation:
Given g (x) = [tex]\frac{x+1}{x-2}[/tex] and [tex]h(x) = 4-x[/tex], we are to find [tex](goh)(-3)[/tex]
First we need to get [tex](goh)(x)[/tex]
[tex](goh)(x) = g(h(x))\\g(h(x))= g(4-x)\\g(4-x) = \frac{(4-x)+1}{(4-x)-2}\\ g(4-x) = \frac{5-x}{2-x}\\substitute \ x = -3 \ into \ resulting \ function\\ g(4-x) = \frac{5-x}{2-x}\\(goh)(-3) = \frac{5-(-3)}{2-(-3)}\\\\(goh)(-3) = \frac{8}{5}\\[/tex]
Hence [tex](goh)(x)\ is \ Eight-fifths[/tex]
Also given f(x) = x and g(x) = 1/x, we are to find [tex](fog)(x)[/tex]
[tex](fog)(x) = f(g(x))\\f(g(x)) = f(\frac{1}{x} )\\ since \ f(x) = x^2, we\ will \ repalce\ x \ with \ \frac{1}{x} \ to \ have;\\ f(\frac{1}{x} ) =( \frac{1}{x})^2\\\\[/tex]
[tex]f(\frac{1}{x} ) = \frac{1}{x^2}[/tex]
For the pair of function f(x) = 2/x and g(x) = 2/x
f(g(x)) = f(2/x)
f(2/x) = 2/(2/x)
f(2/x) = 2*x/2
f(2/x) = x
Hence f(g(x)) = x
For the pair of function f(x) = x-2/3 and g(x) = 2-3x
f(g(x)) = f(2-3x)
f(2-3x) = (2-3x-2)/3
f(2-3x) = -3x/3
f(2-3x) = -x
f(g(x)) = -x for the pair of function
For the pair of function f(x) = x/2 - 2 and g(x) = x/2 + 2
f(g(x)) = f(x/2 + 2)
f(x/2 + 2) = f((x+4)/2)
f((x+4)/2) = [(x+4)/2]/2 - 2
f((x+4)/2) = (x+4)/4 - 2
find the LCM
f((x+4)/2) = [(x+4)-8]/4
f((x+4)/2) = (x-4)/4
Hence f(g(x)) for the pair of function is (x-4)/4
Answer:
The range of g(x) is y > 0
The ranges of f(x) and h(x) are different from the range of g(x)
