Question
consider the functions g(x)=(4-x^2)^0.5 and h(x)=(x^2)-5.
find the composite functions hog(x) and goh(x), if they exist, and state the domain and range of the composite functions.
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ok, i have figured out that hog(x)=-1-X^2, but in my answer booklet it says that the domain of this function is -2<&=x<&=2, and the range is -5<&=y<&=-1. can someone please explain how i can get this answer??thanks
and i have no idea why the function goh(x) does not exist. help please!
find the composite functions hog(x) and goh(x), if they exist, and state the domain and range of the composite functions.
------------------------------
ok, i have figured out that hog(x)=-1-X^2, but in my answer booklet it says that the domain of this function is -2<&=x<&=2, and the range is -5<&=y<&=-1. can someone please explain how i can get this answer??thanks
and i have no idea why the function goh(x) does not exist. help please!
Answers
hog(x) means that you first perform the function g(x), and then enter the result of that function as x in the second function h(x).
The function g(x) is a square root, meaning that whatever is under the square root sign has to be positive (you can't take the square root of a negative number).
4-x^2 is only positive if -2 <= x <= 2, so this is the domain of g(x) and since x can take on any value in h(x), it's also the domain of hog(x).
Now, for the range, we have to check when h(x) takes on its largest and smallest values. When we draw the graph of x²-5, we notice that it is a parabole. We can easily determine that for x=0, the parabole will reach its lowest point, and then starts to rise again. Since the maximum value for x is 2 or -2, we check that for 2 or -2 h(x) reaches the value -1 (being its largest value for -2 <=x <= 2).
So the range of hog(x) is -5<=y<=-1.
As for goh(x), it does exist. if we replace every x in g(x) by h(x), we get goh(x). Since 4-(x²-5)² is only positive for -sqrt(7) until -sqrt(3) and from sqrt(3) untill sqrt(7), this function only has values in those places.
So the domain of goh(x) = [-sqrt(7),-sqrt(3)] U [sqrt(3),sqrt(7)].
The function g(x) is a square root, meaning that whatever is under the square root sign has to be positive (you can't take the square root of a negative number).
4-x^2 is only positive if -2 <= x <= 2, so this is the domain of g(x) and since x can take on any value in h(x), it's also the domain of hog(x).
Now, for the range, we have to check when h(x) takes on its largest and smallest values. When we draw the graph of x²-5, we notice that it is a parabole. We can easily determine that for x=0, the parabole will reach its lowest point, and then starts to rise again. Since the maximum value for x is 2 or -2, we check that for 2 or -2 h(x) reaches the value -1 (being its largest value for -2 <=x <= 2).
So the range of hog(x) is -5<=y<=-1.
As for goh(x), it does exist. if we replace every x in g(x) by h(x), we get goh(x). Since 4-(x²-5)² is only positive for -sqrt(7) until -sqrt(3) and from sqrt(3) untill sqrt(7), this function only has values in those places.
So the domain of goh(x) = [-sqrt(7),-sqrt(3)] U [sqrt(3),sqrt(7)].
write h(x) as composition of two functions h(x) = (fog)(x)
h(x) = 3 square root 7 + 3 square root of x
h(x) = 3 square root 7 + 3 square root of x
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