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

Answer 1

To find the compositions of ( f(x) ) and ( g(x) ), as well as the domain and range:

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

  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}} ]

  3. 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) )

  4. 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) )

  5. 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) )

  6. 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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Answer 2

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.

I will show you how to do #ƒ(g(x))#. This means that we must plug in function g in for x in function ƒ.
#ƒ(g(x)) = ƒ(x - 1)#
#= 1/(x - 1)^2#
#=1/(x^2 - x - x + 1)#
#=1/(x^2 - 2x + 1)#
So, #ƒ(g(x)) = 1/(x^2 - 2x + 1)#

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]}

When a number replaces x in the parentheses (e.g: #ƒ(g(x)) -> ƒ(g(2))#), you must first plug the value in the parentheses into the inner function (g in this case), and then plug in the answer of that calculation into the outer function (ƒ in this case). If you have more than 2 functions, do the same thing, but just remember to work from the inside to the outside.

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.

#x^2 - 2x + 1 = 0#
#(x - 1)(x - 1) = 0#
#x = 1#
The domain would be #x in RR# | #x != 1#.

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.

So, the range is #y > 0#

Practice exercises:

a) #g(h(x))#
b) #h(ƒ(x))#
c) #g(ƒ(8))#
d) #ƒ(g(h(x)))#

Good luck!

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Answer 3

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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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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