Let #f(x) = 1/x^2# and #g(x) = x−1#, how do you find each of the compositions and domain and range?
To find the compositions of ( f(x) ) and ( g(x) ), as well as the domain and range:

Composition of ( f(g(x)) ): [ f(g(x)) = f(x  1) = \frac{1}{{(x  1)^2}} ]

Composition of ( g(f(x)) ): [ g(f(x)) = g\left(\frac{1}{{x^2}}\right) = \frac{1}{{x^2}}  1 = \frac{1  x^2}{{x^2}} ]

Domain of ( f(g(x)) ): All real numbers except ( x = 1 ) since ( (x  1)^2 ) cannot be zero. Domain: ( x \in (\infty, 1) \cup (1, \infty) )

Range of ( f(g(x)) ): Since ( f(x) = 1/x^2 ), the range of ( f(g(x)) ) is all real numbers except ( 0 ). Range: ( y \in (\infty, 0) \cup (0, \infty) )

Domain of ( g(f(x)) ): All real numbers except ( x = 0 ) since ( x^2 ) cannot be zero. Domain: ( x \in (\infty, 0) \cup (0, \infty) )

Range of ( g(f(x)) ): Since ( g(x) = x  1 ), the range of ( g(f(x)) ) is all real numbers. Range: ( y \in (\infty, \infty) )
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There are many compositions possible. I will show you how to do one. See practice exercises at bottom of page for you to practice the new found skill.
As for the domain and range, the domain are all permissible values of x and the range is the same, except for the values of y.
Here is the graph of the new function.
graph{1/(x^2  2x + 1) [10, 10, 5, 5]}
Since having the denominator equal to 0 in a rational or reciprocal function is undefined (because division by 0 is undefined), we must set the denominator to 0 and solve for x.
Range: Looking at the graph, you can see, as you zoom out, that the graph becomes extremely close to, but never touches, the horizontal line y = 0. Also, you can see that the function never goes below the line y = 0. In other words, the y value must always stay positive.
Practice exercises:
Good luck!
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To find the composition of ( f ) and ( g ), and ( g ) and ( f ), first, express the compositions as ( (f \circ g)(x) ) and ( (g \circ f)(x) ), respectively. Then, substitute the functions into each other, considering the inner function as the argument of the outer function.
( (f \circ g)(x) = f(g(x)) = f(x  1) = \frac{1}{(x  1)^2} )
( (g \circ f)(x) = g(f(x)) = g(\frac{1}{x^2}) = \frac{1}{x^2}  1 )
For the domain of each composition, consider the restrictions imposed by the functions involved. For ( (f \circ g)(x) ), the denominator cannot be zero, so ( x  1 \neq 0 ), which implies ( x \neq 1 ). For ( (g \circ f)(x) ), the denominator ( x^2 ) cannot be zero, so ( x \neq 0 ). Therefore, the domains are:
( (f \circ g)(x) ) has the domain ( x \in (\infty, 1) \cup (1, \infty) )
( (g \circ f)(x) ) has the domain ( x \in (\infty, 0) \cup (0, \infty) )
The range of each composition depends on the behavior of the function and the restrictions of the domain. Since ( f(x) = \frac{1}{x^2} ) and ( g(x) = x  1 ), the range of ( (f \circ g)(x) ) is all real numbers except 0, and the range of ( (g \circ f)(x) ) is all real numbers.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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