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Sagot :
Alright, let's go through each part of the question step by step.
### Part a: What is [tex]\( f(4) \)[/tex]?
To find [tex]\( f(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into the function [tex]\( f(x) = 3x^2 - 5 \)[/tex]:
[tex]\[ f(4) = 3(4)^2 - 5 \][/tex]
After calculating,
[tex]\[ f(4) = 3(16) - 5 \][/tex]
[tex]\[ f(4) = 48 - 5 \][/tex]
[tex]\[ f(4) = 43 \][/tex]
Hence, [tex]\( f(4) = 43 \)[/tex].
### Part b: What is [tex]\( g(4) \)[/tex]?
To find [tex]\( g(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into the function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex]:
[tex]\[ g(4) = \sqrt{4 - 3} + 2 \][/tex]
After calculating,
[tex]\[ g(4) = \sqrt{1} + 2 \][/tex]
[tex]\[ g(4) = 1 + 2 \][/tex]
[tex]\[ g(4) = 3 \][/tex]
Hence, [tex]\( g(4) = 3 \)[/tex].
### Part c: What is [tex]\( f(x) \)[/tex] if [tex]\( x=7 \)[/tex]?
To find [tex]\( f(7) \)[/tex], substitute [tex]\( x = 7 \)[/tex] into the function [tex]\( f(x) = 3x^2 - 5 \)[/tex]:
[tex]\[ f(7) = 3(7)^2 - 5 \][/tex]
After calculating,
[tex]\[ f(7) = 3(49) - 5 \][/tex]
[tex]\[ f(7) = 147 - 5 \][/tex]
[tex]\[ f(7) = 142 \][/tex]
Hence, [tex]\( f(7) = 142 \)[/tex].
### Part d: What is [tex]\( g(x) \)[/tex] if [tex]\( x=3 \)[/tex]?
To find [tex]\( g(3) \)[/tex], substitute [tex]\( x = 3 \)[/tex] into the function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex]:
[tex]\[ g(3) = \sqrt{3 - 3} + 2 \][/tex]
After calculating,
[tex]\[ g(3) = \sqrt{0} + 2 \][/tex]
[tex]\[ g(3) = 0 + 2 \][/tex]
[tex]\[ g(3) = 2 \][/tex]
Hence, [tex]\( g(3) = 2 \)[/tex].
### Part e: Describe the domain of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex].
- The function [tex]\( f(x) = 3x^2 - 5 \)[/tex] is a quadratic function. Quadratic functions are defined for all real numbers. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f(x) = (-\infty, \infty) \][/tex]
- The function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex] involves a square root. The expression inside the square root must be non-negative for the function to be defined. Hence, [tex]\( x - 3 \geq 0 \)[/tex] or [tex]\( x \geq 3 \)[/tex]. Therefore, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Domain of } g(x) = [3, \infty) \][/tex]
### Part f: Why is the domain of one of these functions more restrictive than the other?
The domain of [tex]\( g(x) \)[/tex] is more restrictive than the domain of [tex]\( f(x) \)[/tex] because the square root function [tex]\( \sqrt{x - 3} \)[/tex] requires the expression [tex]\( x - 3 \)[/tex] to be non-negative to produce real numbers. This restriction does not apply to [tex]\( f(x) \)[/tex], which is a polynomial and therefore defined for all real numbers. Thus, [tex]\( g(x) \)[/tex] has a more limited domain, starting at [tex]\( x = 3 \)[/tex] and extending to infinity.
### Part a: What is [tex]\( f(4) \)[/tex]?
To find [tex]\( f(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into the function [tex]\( f(x) = 3x^2 - 5 \)[/tex]:
[tex]\[ f(4) = 3(4)^2 - 5 \][/tex]
After calculating,
[tex]\[ f(4) = 3(16) - 5 \][/tex]
[tex]\[ f(4) = 48 - 5 \][/tex]
[tex]\[ f(4) = 43 \][/tex]
Hence, [tex]\( f(4) = 43 \)[/tex].
### Part b: What is [tex]\( g(4) \)[/tex]?
To find [tex]\( g(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into the function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex]:
[tex]\[ g(4) = \sqrt{4 - 3} + 2 \][/tex]
After calculating,
[tex]\[ g(4) = \sqrt{1} + 2 \][/tex]
[tex]\[ g(4) = 1 + 2 \][/tex]
[tex]\[ g(4) = 3 \][/tex]
Hence, [tex]\( g(4) = 3 \)[/tex].
### Part c: What is [tex]\( f(x) \)[/tex] if [tex]\( x=7 \)[/tex]?
To find [tex]\( f(7) \)[/tex], substitute [tex]\( x = 7 \)[/tex] into the function [tex]\( f(x) = 3x^2 - 5 \)[/tex]:
[tex]\[ f(7) = 3(7)^2 - 5 \][/tex]
After calculating,
[tex]\[ f(7) = 3(49) - 5 \][/tex]
[tex]\[ f(7) = 147 - 5 \][/tex]
[tex]\[ f(7) = 142 \][/tex]
Hence, [tex]\( f(7) = 142 \)[/tex].
### Part d: What is [tex]\( g(x) \)[/tex] if [tex]\( x=3 \)[/tex]?
To find [tex]\( g(3) \)[/tex], substitute [tex]\( x = 3 \)[/tex] into the function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex]:
[tex]\[ g(3) = \sqrt{3 - 3} + 2 \][/tex]
After calculating,
[tex]\[ g(3) = \sqrt{0} + 2 \][/tex]
[tex]\[ g(3) = 0 + 2 \][/tex]
[tex]\[ g(3) = 2 \][/tex]
Hence, [tex]\( g(3) = 2 \)[/tex].
### Part e: Describe the domain of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex].
- The function [tex]\( f(x) = 3x^2 - 5 \)[/tex] is a quadratic function. Quadratic functions are defined for all real numbers. Therefore, the domain of [tex]\( f(x) \)[/tex] is all real numbers:
[tex]\[ \text{Domain of } f(x) = (-\infty, \infty) \][/tex]
- The function [tex]\( g(x) = \sqrt{x - 3} + 2 \)[/tex] involves a square root. The expression inside the square root must be non-negative for the function to be defined. Hence, [tex]\( x - 3 \geq 0 \)[/tex] or [tex]\( x \geq 3 \)[/tex]. Therefore, the domain of [tex]\( g(x) \)[/tex] is:
[tex]\[ \text{Domain of } g(x) = [3, \infty) \][/tex]
### Part f: Why is the domain of one of these functions more restrictive than the other?
The domain of [tex]\( g(x) \)[/tex] is more restrictive than the domain of [tex]\( f(x) \)[/tex] because the square root function [tex]\( \sqrt{x - 3} \)[/tex] requires the expression [tex]\( x - 3 \)[/tex] to be non-negative to produce real numbers. This restriction does not apply to [tex]\( f(x) \)[/tex], which is a polynomial and therefore defined for all real numbers. Thus, [tex]\( g(x) \)[/tex] has a more limited domain, starting at [tex]\( x = 3 \)[/tex] and extending to infinity.
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