Question

If g (x) = StartFraction x + 1 Over x minus 2 EndFraction and h(x) = 4 – x, what is the value of (g circle h) (negative 3)? Eight-fifths Five-halves Fifteen-halves Eighteen-fifthsFor which pairs of functions is (f circle g) (x)? f (x) = x squared and g (x) = StartFraction 1 Over x EndFraction f (x) = StartFraction 2 Over x EndFraction and g (x) = StartFraction 2 Over x EndFraction f (x) = StartFraction x minus 2 Over 3 EndFraction and g (x) = 2 minus 3 x f (x) = one-half x minus 2 and g (x) = one-half x + 2

Answer 1

The range of g(x) is y > 0

The ranges of f(x) and h(x) are different from the range of g(x)

Answer 2

Explanation:

Given g (x) =
`(x+1)/(x-2)` and
`h(x) = 4-x`, we are to find
`(goh)(-3)`

First we need to get
`(goh)(x)`

`(goh)(x) = g(h(x))\\g(h(x))= g(4-x)\\g(4-x) = ((4-x)+1)/((4-x)-2)\\ g(4-x) = (5-x)/(2-x)\\substitute \ x = -3 \ into \ resulting \ function\\ g(4-x) = (5-x)/(2-x)\\(goh)(-3) = (5-(-3))/(2-(-3))\\\\(goh)(-3) = (8)/(5)\\`

Hence
`(goh)(x)\ is \ Eight-fifths`

Also given f(x) = x and g(x) = 1/x, we are to find
`(fog)(x)`

`(fog)(x) = f(g(x))\\f(g(x)) = f((1)/(x) )\\ since \ f(x) = x^2, we\ will \ repalce\ x \ with \ (1)/(x) \ to \ have;\\ f((1)/(x) ) =( (1)/(x))^2\\\\`

`f((1)/(x) ) = (1)/(x^2)`

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