Answer :
Sure! Let's consider the given functions and their compositions:
1. [tex]\( f(x) = \frac{1}{\sqrt{x}} \)[/tex]
2. [tex]\( g(x) = x^2 - 4x \)[/tex]
### Composition [tex]\( f \circ g \)[/tex]
To find [tex]\( f \circ g \)[/tex], we need:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
This means we first apply [tex]\( g(x) \)[/tex] and then apply [tex]\( f \)[/tex] to the result:
[tex]\[ g(x) = x^2 - 4x \][/tex]
Next, we apply [tex]\( f \)[/tex] to this:
[tex]\[ f(g(x)) = f(x^2 - 4x) = \frac{1}{\sqrt{x^2 - 4x}} \][/tex]
Now, we need the domain of [tex]\( f \circ g \)[/tex]. For [tex]\( f(x) \)[/tex], the expression inside the square root must be positive:
[tex]\[ x^2 - 4x > 0 \][/tex]
Solving this inequality, we find the roots:
[tex]\[ x(x - 4) > 0 \][/tex]
The solution is the intervals where the product is positive:
[tex]\[ (-\infty, 0) \cup (4, \infty) \][/tex]
So, the domain of [tex]\( f \circ g \)[/tex] is:
[tex]\[ (-\infty, 0) \cup (4, \infty) \][/tex]
### Composition [tex]\( g \circ f \)[/tex]
To find [tex]\( g \circ f \)[/tex], we need:
[tex]\[ (g \circ f)(x) = g(f(x)) \][/tex]
This means we first apply [tex]\( f(x) \)[/tex] and then apply [tex]\( g \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
Next, we apply [tex]\( g \)[/tex] to this:
[tex]\[ g(f(x)) = g\left( \frac{1}{\sqrt{x}} \right) = \left( \frac{1}{\sqrt{x}} \right)^2 - 4 \left( \frac{1}{\sqrt{x}} \right) \][/tex]
[tex]\[ = \frac{1}{x} - \frac{4}{\sqrt{x}} \][/tex]
Now, we need the domain of [tex]\( g \circ f \)[/tex]. [tex]\( f(x) \)[/tex] is defined for [tex]\( x > 0 \)[/tex]. Since [tex]\( g \)[/tex] is always defined for [tex]\( g(x) \)[/tex]:
[tex]\[ x > 0 \][/tex]
So, the domain of [tex]\( g \circ f \)[/tex] is:
[tex]\[ (0, \infty) \][/tex]
### Composition [tex]\( f \circ f \)[/tex]
To find [tex]\( f \circ f \)[/tex], we need:
[tex]\[ (f \circ f)(x) = f(f(x)) \][/tex]
This means we first apply [tex]\( f(x) \)[/tex] and then apply [tex]\( f \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
Next, we apply [tex]\( f \)[/tex] to this:
[tex]\[ f(f(x)) = f\left( \frac{1}{\sqrt{x}} \right) = \frac{1}{\sqrt{\frac{1}{\sqrt{x}}}} = \frac{1}{\left( \frac{1}{x^{1/4}} \right)} \][/tex]
[tex]\[ = x^{1/4} \][/tex]
Now, we need the domain for [tex]\( f \circ f \)[/tex]:
Since [tex]\( f(x) \)[/tex] is defined for [tex]\( x > 0 \)[/tex], [tex]\( f(f(x)) \)[/tex] requires [tex]\( x > 0 \)[/tex].
So, the domain of [tex]\( f \circ f \)[/tex] is:
[tex]\[ (0, \infty) \][/tex]
### Composition [tex]\( g \circ g \)[/tex]
To find [tex]\( g \circ g \)[/tex], we need:
[tex]\[ (g \circ g)(x) = g(g(x)) \][/tex]
This means we first apply [tex]\( g(x) \)[/tex] and then apply [tex]\( g \)[/tex]:
[tex]\[ g(x) = x^2 - 4x \][/tex]
Next, we apply [tex]\( g \)[/tex] to this:
[tex]\[ g(g(x)) = g(x^2 - 4x) = (x^2 - 4x)^2 - 4(x^2 - 4x) \][/tex]
[tex]\[ = x^4 - 8x^3 + 16x^2 - 4x^2 + 16x \][/tex]
[tex]\[ = x^4 - 8x^3 + 12x^2 + 16x \][/tex]
Now, we need the domain of [tex]\( g \circ g \)[/tex]:
Since [tex]\( g(x) \)[/tex] is defined for all real numbers [tex]\( x \)[/tex], [tex]\( g \circ g \)[/tex] is also defined for all real numbers.
So, the domain of [tex]\( g \circ g \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
### Summary
- [tex]\( f \circ g(x) = \frac{1}{\sqrt{x^2 - 4x}} \)[/tex] with domain [tex]\( (-\infty, 0) \cup (4, \infty) \)[/tex]
- [tex]\( g \circ f(x) = \frac{1}{x} - \frac{4}{\sqrt{x}} \)[/tex] with domain [tex]\( (0, \infty) \)[/tex]
- [tex]\( f \circ f(x) = x^{1/4} \)[/tex] with domain [tex]\( (0, \infty) \)[/tex]
- [tex]\( g \circ g(x) = x^4 - 8x^3 + 12x^2 + 16x \)[/tex] with domain [tex]\( (-\infty, \infty) \)[/tex]
1. [tex]\( f(x) = \frac{1}{\sqrt{x}} \)[/tex]
2. [tex]\( g(x) = x^2 - 4x \)[/tex]
### Composition [tex]\( f \circ g \)[/tex]
To find [tex]\( f \circ g \)[/tex], we need:
[tex]\[ (f \circ g)(x) = f(g(x)) \][/tex]
This means we first apply [tex]\( g(x) \)[/tex] and then apply [tex]\( f \)[/tex] to the result:
[tex]\[ g(x) = x^2 - 4x \][/tex]
Next, we apply [tex]\( f \)[/tex] to this:
[tex]\[ f(g(x)) = f(x^2 - 4x) = \frac{1}{\sqrt{x^2 - 4x}} \][/tex]
Now, we need the domain of [tex]\( f \circ g \)[/tex]. For [tex]\( f(x) \)[/tex], the expression inside the square root must be positive:
[tex]\[ x^2 - 4x > 0 \][/tex]
Solving this inequality, we find the roots:
[tex]\[ x(x - 4) > 0 \][/tex]
The solution is the intervals where the product is positive:
[tex]\[ (-\infty, 0) \cup (4, \infty) \][/tex]
So, the domain of [tex]\( f \circ g \)[/tex] is:
[tex]\[ (-\infty, 0) \cup (4, \infty) \][/tex]
### Composition [tex]\( g \circ f \)[/tex]
To find [tex]\( g \circ f \)[/tex], we need:
[tex]\[ (g \circ f)(x) = g(f(x)) \][/tex]
This means we first apply [tex]\( f(x) \)[/tex] and then apply [tex]\( g \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
Next, we apply [tex]\( g \)[/tex] to this:
[tex]\[ g(f(x)) = g\left( \frac{1}{\sqrt{x}} \right) = \left( \frac{1}{\sqrt{x}} \right)^2 - 4 \left( \frac{1}{\sqrt{x}} \right) \][/tex]
[tex]\[ = \frac{1}{x} - \frac{4}{\sqrt{x}} \][/tex]
Now, we need the domain of [tex]\( g \circ f \)[/tex]. [tex]\( f(x) \)[/tex] is defined for [tex]\( x > 0 \)[/tex]. Since [tex]\( g \)[/tex] is always defined for [tex]\( g(x) \)[/tex]:
[tex]\[ x > 0 \][/tex]
So, the domain of [tex]\( g \circ f \)[/tex] is:
[tex]\[ (0, \infty) \][/tex]
### Composition [tex]\( f \circ f \)[/tex]
To find [tex]\( f \circ f \)[/tex], we need:
[tex]\[ (f \circ f)(x) = f(f(x)) \][/tex]
This means we first apply [tex]\( f(x) \)[/tex] and then apply [tex]\( f \)[/tex]:
[tex]\[ f(x) = \frac{1}{\sqrt{x}} \][/tex]
Next, we apply [tex]\( f \)[/tex] to this:
[tex]\[ f(f(x)) = f\left( \frac{1}{\sqrt{x}} \right) = \frac{1}{\sqrt{\frac{1}{\sqrt{x}}}} = \frac{1}{\left( \frac{1}{x^{1/4}} \right)} \][/tex]
[tex]\[ = x^{1/4} \][/tex]
Now, we need the domain for [tex]\( f \circ f \)[/tex]:
Since [tex]\( f(x) \)[/tex] is defined for [tex]\( x > 0 \)[/tex], [tex]\( f(f(x)) \)[/tex] requires [tex]\( x > 0 \)[/tex].
So, the domain of [tex]\( f \circ f \)[/tex] is:
[tex]\[ (0, \infty) \][/tex]
### Composition [tex]\( g \circ g \)[/tex]
To find [tex]\( g \circ g \)[/tex], we need:
[tex]\[ (g \circ g)(x) = g(g(x)) \][/tex]
This means we first apply [tex]\( g(x) \)[/tex] and then apply [tex]\( g \)[/tex]:
[tex]\[ g(x) = x^2 - 4x \][/tex]
Next, we apply [tex]\( g \)[/tex] to this:
[tex]\[ g(g(x)) = g(x^2 - 4x) = (x^2 - 4x)^2 - 4(x^2 - 4x) \][/tex]
[tex]\[ = x^4 - 8x^3 + 16x^2 - 4x^2 + 16x \][/tex]
[tex]\[ = x^4 - 8x^3 + 12x^2 + 16x \][/tex]
Now, we need the domain of [tex]\( g \circ g \)[/tex]:
Since [tex]\( g(x) \)[/tex] is defined for all real numbers [tex]\( x \)[/tex], [tex]\( g \circ g \)[/tex] is also defined for all real numbers.
So, the domain of [tex]\( g \circ g \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
### Summary
- [tex]\( f \circ g(x) = \frac{1}{\sqrt{x^2 - 4x}} \)[/tex] with domain [tex]\( (-\infty, 0) \cup (4, \infty) \)[/tex]
- [tex]\( g \circ f(x) = \frac{1}{x} - \frac{4}{\sqrt{x}} \)[/tex] with domain [tex]\( (0, \infty) \)[/tex]
- [tex]\( f \circ f(x) = x^{1/4} \)[/tex] with domain [tex]\( (0, \infty) \)[/tex]
- [tex]\( g \circ g(x) = x^4 - 8x^3 + 12x^2 + 16x \)[/tex] with domain [tex]\( (-\infty, \infty) \)[/tex]
