Discover new perspectives and gain insights with IDNLearn.com. Ask anything and receive well-informed answers from our community of experienced professionals.
Sagot :
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]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.
