This is a composite functions, since the given are


f(x) =x^2-x-6 g(x)= the square root of √x/x
h(x)=g(f(x))

then

h(x) = √(x^2-x-6) / (x^2-x-6) or h(x) = 1 / √(x^2-x-6)

To find the domain of h(x), since h(x) is a combination of radical function and rational function, we need to find the zeroes of the function inside the radical that is , find the zeroes of f(x)

using factoring f(x) = x^2-x-6

x^2-x-6 = 0

(x - 3)(x + 2) = 0

x = 3 or x = -2

so the possible regions are -2 < x 3 or x 3 or x 3 or x < -2

using Numberline, the domain will be






-5 -4 -3 -2 -1 0 1 2 3 4 5


(note again: please disregard the dots above the interval between -2 and 3, since there should not be any dots there)

To know if it is an open or close interval, try substituting a value between the two boundaries, so between -2 and 3, try substituting any value, let us say, zero, if the value of h(x) is real number, then it is a closed interval, otherwise, it is an open interval


Similarly if the interval is open, then the arrows should be away, directed to left and right (to + and - infinity) but if the interval is closed, there are no arrows)



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Here are the answers



http://www.mediafire.com/view/?yftofl8zdxkzyd4




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ah ok, here are the step by step solutions


if(x) =x^2-x-6 g(x)= √x/x
h(x)=g(f(x))


example: f(x) = 2x + 1, g(x) = √x, and h(x) = g(f(x))

then since f(x) is inside g(x), we put f(x) = 2x + 1, inside g(x), thus

h(x) = g(f(x))

h(x) = g(2x + 1)

h(x) = √(2x + 1)

similarly


if(x) = x^2-x-6 g(x)= √x/x
h(x)=g(f(x))


then putting or substituting x^2-x-6 for x in g(x)= √x/x, then

g(x)= √x/x --> g(x^2-x-6)= √(x^2-x-6)/(x^2-x-6)


thus h(x) = √(x^2-x-6)/(x^2-x-6)


(note: that √2/2 is also the same as 1 / √2, similarly, √5/5 = 1/√5)

and so √(x^2-x-6)/(x^2-x-6) = 1/ √(x^2-x-6)



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here they are so the possible regions are x < 3 and x > -2