Answer :
To compare and contrast the piecewise defined functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], we will examine their behavior in different segments of the domain.
### Definitions:
[tex]\[ f(x)=\left\{ \begin{array}{rl} -x + 2, & x < 0 \\ x^2 + 1, & x > 0 \end{array} \right. \][/tex]
[tex]\[ g(x)=\left\{ \begin{array}{ll} x + 2, & x < 0 \\ x^2 + 2, & x > 0 \end{array} \right. \][/tex]
### Comparisons at Specific Points:
We will evaluate and compare [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at the points [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = 2 \)[/tex].
1. At [tex]\( x = -2 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = -(-2) + 2 = 2 + 2 = 4 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2) = -2 + 2 = 0 \][/tex]
Thus, at [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4 \][/tex]
[tex]\[ g(-2) = 0 \][/tex]
2. At [tex]\( x = -1 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(-1) = -(-1) + 2 = 1 + 2 = 3 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(-1) = -1 + 2 = 1 \][/tex]
Thus, at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3 \][/tex]
[tex]\[ g(-1) = 1 \][/tex]
3. At [tex]\( x = 1 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(1) = 1^2 + 1 = 1 + 1 = 2 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(1) = 1^2 + 2 = 1 + 2 = 3 \][/tex]
Thus, at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2 \][/tex]
[tex]\[ g(1) = 3 \][/tex]
4. At [tex]\( x = 2 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(2) = 2^2 + 1 = 4 + 1 = 5 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(2) = 2^2 + 2 = 4 + 2 = 6 \][/tex]
Thus, at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 \][/tex]
[tex]\[ g(2) = 6 \][/tex]
### Summary of Results:
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4, \quad g(-2) = 0 \][/tex]
- At [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3, \quad g(-1) = 1 \][/tex]
- At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2, \quad g(1) = 3 \][/tex]
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5, \quad g(2) = 6 \][/tex]
### Comparison and Contrast:
1. For [tex]\( x < 0 \)[/tex]:
- For negative values, both functions are linear.
- [tex]\( f(x) \)[/tex] decreases as [tex]\( x \)[/tex] decreases (i.e., slope = -1), while [tex]\( g(x) \)[/tex] increases as [tex]\( x \)[/tex] decreases (i.e., slope = 1).
- The values of [tex]\( f(x) \)[/tex] are higher than those of [tex]\( g(x) \)[/tex] for the same negative [tex]\( x \)[/tex].
2. For [tex]\( x > 0 \)[/tex]:
- For positive values, both functions are quadratic.
- [tex]\( f(x) = x^2 + 1 \)[/tex] is always less than [tex]\( g(x) = x^2 + 2 \)[/tex] by exactly 1 for any [tex]\( x > 0 \)[/tex].
### Conclusion:
While both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are piecewise defined functions with linear components for [tex]\( x < 0 \)[/tex] and quadratic components for [tex]\( x > 0 \)[/tex], they differ in both segments of their domain. For [tex]\( x < 0 \)[/tex], [tex]\(f(x)\)[/tex] yields higher values than [tex]\(g(x)\)[/tex]. For [tex]\( x > 0 \)[/tex], [tex]\(g(x)\)[/tex] exceeds [tex]\(f(x)\)[/tex] by a constant difference of 1 unit.
### Definitions:
[tex]\[ f(x)=\left\{ \begin{array}{rl} -x + 2, & x < 0 \\ x^2 + 1, & x > 0 \end{array} \right. \][/tex]
[tex]\[ g(x)=\left\{ \begin{array}{ll} x + 2, & x < 0 \\ x^2 + 2, & x > 0 \end{array} \right. \][/tex]
### Comparisons at Specific Points:
We will evaluate and compare [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] at the points [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], [tex]\( x = 1 \)[/tex], and [tex]\( x = 2 \)[/tex].
1. At [tex]\( x = -2 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(-2) = -(-2) + 2 = 2 + 2 = 4 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(-2) = -2 + 2 = 0 \][/tex]
Thus, at [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4 \][/tex]
[tex]\[ g(-2) = 0 \][/tex]
2. At [tex]\( x = -1 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(-1) = -(-1) + 2 = 1 + 2 = 3 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(-1) = -1 + 2 = 1 \][/tex]
Thus, at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3 \][/tex]
[tex]\[ g(-1) = 1 \][/tex]
3. At [tex]\( x = 1 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(1) = 1^2 + 1 = 1 + 1 = 2 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(1) = 1^2 + 2 = 1 + 2 = 3 \][/tex]
Thus, at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2 \][/tex]
[tex]\[ g(1) = 3 \][/tex]
4. At [tex]\( x = 2 \)[/tex]:
- For [tex]\( f(x) \)[/tex]:
[tex]\[ f(2) = 2^2 + 1 = 4 + 1 = 5 \][/tex]
- For [tex]\( g(x) \)[/tex]:
[tex]\[ g(2) = 2^2 + 2 = 4 + 2 = 6 \][/tex]
Thus, at [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5 \][/tex]
[tex]\[ g(2) = 6 \][/tex]
### Summary of Results:
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 4, \quad g(-2) = 0 \][/tex]
- At [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 3, \quad g(-1) = 1 \][/tex]
- At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 2, \quad g(1) = 3 \][/tex]
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 5, \quad g(2) = 6 \][/tex]
### Comparison and Contrast:
1. For [tex]\( x < 0 \)[/tex]:
- For negative values, both functions are linear.
- [tex]\( f(x) \)[/tex] decreases as [tex]\( x \)[/tex] decreases (i.e., slope = -1), while [tex]\( g(x) \)[/tex] increases as [tex]\( x \)[/tex] decreases (i.e., slope = 1).
- The values of [tex]\( f(x) \)[/tex] are higher than those of [tex]\( g(x) \)[/tex] for the same negative [tex]\( x \)[/tex].
2. For [tex]\( x > 0 \)[/tex]:
- For positive values, both functions are quadratic.
- [tex]\( f(x) = x^2 + 1 \)[/tex] is always less than [tex]\( g(x) = x^2 + 2 \)[/tex] by exactly 1 for any [tex]\( x > 0 \)[/tex].
### Conclusion:
While both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are piecewise defined functions with linear components for [tex]\( x < 0 \)[/tex] and quadratic components for [tex]\( x > 0 \)[/tex], they differ in both segments of their domain. For [tex]\( x < 0 \)[/tex], [tex]\(f(x)\)[/tex] yields higher values than [tex]\(g(x)\)[/tex]. For [tex]\( x > 0 \)[/tex], [tex]\(g(x)\)[/tex] exceeds [tex]\(f(x)\)[/tex] by a constant difference of 1 unit.
