Question 828013: I need finding the composition of functions (f o g)(x).?
1. g(x) = 3/(x - 1), f(x) = (x - 1)/(x - 3)
A. (x - 1)/(x - 3) * 3/(x - 1)
B. (x - 1)/(x - 1) * 3/(x - 1)
C. (x - 4)/(x - 2)
D. (x - 4)/(3x - 3)
E. (x - 4)/(3x - 6)
F. 3x^2
2. g(x) = x2 - 8, f(x) = (-x + 1)^1/2
A. (-x2 + 9)^(1/2)
B. (-x2 - 8)^(1/2)
C. (-x + 1)^(1/2) (x2 - 8)
D. (-x + 1)^(1/2) (x2 - 8)
E. (-x2 - 9)^(1/2)
F. (x - 2)^2
3. g(x) = (x2 -16)^(1/2), f(x) = (3 - x)^(1/2)
A. (3 - x)^(1/2), (x2 -16)^(1/2)
B. (-x-13)^(1/2)
C. (3-(x2 -16)^(1/4))
D. (3x2 + 16)
E. (3x(1/2) + 16)
F. (3-(x2 -16)^(1/2))^(1/2) Answer by jsmallt9(3758) (Show Source):
The expression between the parentheses represents the input to the function. Here the input is an x. f(4) on the other hand indicates that the input is a 4.
The whole expression, "f(x)" represents the output of the function (when the input is x).
(f o g)(x) is another way of writing f(g(x)). In the parentheses for the function f we see "g(x)". This means that we will be using g(x) as input to function f:
(f o g)(x)
f(g(x))
Substituting g(x) in we get:
Using as input to f we get:
This is not one of the provided answers. So we need to start simplifying until we get a match. To eliminate the "little" fractions within the "big" fraction we multiply the numerator and denominator of the "big" fraction by the lowest common denominator (LCD) of the "little" denominators. Since both "little" denominators are (x-1), the LCD is (x-1):
To multiply we will need to use the Distributive Property:
Simplifying...
This is still not one of the given answers. But it is close to answer E. In fact the our numerator, 4-x, and E's numerator, x-4, are opposites of each other. And so are the two denominators! So if we multiply the numerator and denominator by -1:
and we have answer E!
I'll leave the other two for you to finish. (Please do not include so many problems, especially ones that take substantial effort, in a single post!)
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