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

Answer:

Step-by-step explanation:

Given  g (x) = $\frac{x+1}{x-2}$ and $h(x) = 4-x$, we are to find $(goh)(-3)$

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

$(goh)(x) = g(h(x))\\g(h(x))= g(4-x)\\g(4-x) = \frac{(4-x)+1}{(4-x)-2}\\ g(4-x) = \frac{5-x}{2-x}\\substitute \ x = -3 \ into \ resulting \ function\\ g(4-x) = \frac{5-x}{2-x}\\(goh)(-3) = \frac{5-(-3)}{2-(-3)}\\\\(goh)(-3) = \frac{8}{5}$

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

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

$(fog)(x) = f(g(x))\\f(g(x)) = f(\frac{1}{x} )\\ since \ f(x) = x^2, we\ will \ repalce\ x \ with \ \frac{1}{x} \ to \ have;\\ f(\frac{1}{x} ) =( \frac{1}{x})^2$

$f(\frac{1}{x} ) = \frac{1}{x^2}$

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