Answer :
Sure, let's work through this step-by-step.
Given:
- The domain of [tex]\( y = g(x) \)[/tex] is [tex]\( -2 \leq x \leq 11 \)[/tex].
- The range of [tex]\( y = g(x) \)[/tex] is [tex]\( -16 \leq y \leq 8 \)[/tex].
(a) For the function [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex]:
Domain:
Since [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex] implies that [tex]\( x \)[/tex] is replaced with [tex]\( \frac{1}{3} x \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of the original domain of [tex]\( g \)[/tex]:
[tex]\[ -2 \leq \frac{1}{3} x \leq 11 \][/tex]
To clear the fraction, multiply through by 3:
[tex]\[ -2 \cdot 3 \leq x \leq 11 \cdot 3 \][/tex]
[tex]\[ -6 \leq x \leq 33 \][/tex]
So, the domain is:
[tex]\[ -6 \leq x \leq 33 \][/tex]
Range:
The function [tex]\( g\left(\frac{1}{3} x\right) \)[/tex] does not change the range of [tex]\( g(x) \)[/tex], because only the input [tex]\( x \)[/tex] is being scaled, not the output. Thus, the range remains:
[tex]\[ -16 \leq y \leq 8 \][/tex]
(b) For the function [tex]\( y = 4g(x) \)[/tex]:
Domain:
The domain of [tex]\( y = 4g(x) \)[/tex] is the same as the original domain of [tex]\( g(x) \)[/tex], since the transformation involves only scaling the output, not the input. Therefore, the domain is:
[tex]\[ -2 \leq x \leq 11 \][/tex]
Range:
Since we are multiplying the output of [tex]\( g(x) \)[/tex] by 4, we need to scale the original range:
[tex]\[ y = 4 \cdot (-16) \quad \text{to} \quad y = 4 \cdot 8 \][/tex]
[tex]\[ -64 \leq y \leq 32 \][/tex]
So, the range is:
[tex]\[ -64 \leq y \leq 32 \][/tex]
Summarizing, the solutions for the domains and ranges are:
(a) For [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex]:
- Domain: [tex]\( -6 \leq x \leq 33 \)[/tex]
- Range: [tex]\( -16 \leq y \leq 8 \)[/tex]
(b) For [tex]\( y = 4g(x) \)[/tex]:
- Domain: [tex]\( -2 \leq x \leq 11 \)[/tex]
- Range: [tex]\( -64 \leq y \leq 32 \)[/tex]
Given:
- The domain of [tex]\( y = g(x) \)[/tex] is [tex]\( -2 \leq x \leq 11 \)[/tex].
- The range of [tex]\( y = g(x) \)[/tex] is [tex]\( -16 \leq y \leq 8 \)[/tex].
(a) For the function [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex]:
Domain:
Since [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex] implies that [tex]\( x \)[/tex] is replaced with [tex]\( \frac{1}{3} x \)[/tex], we need to solve for [tex]\( x \)[/tex] in terms of the original domain of [tex]\( g \)[/tex]:
[tex]\[ -2 \leq \frac{1}{3} x \leq 11 \][/tex]
To clear the fraction, multiply through by 3:
[tex]\[ -2 \cdot 3 \leq x \leq 11 \cdot 3 \][/tex]
[tex]\[ -6 \leq x \leq 33 \][/tex]
So, the domain is:
[tex]\[ -6 \leq x \leq 33 \][/tex]
Range:
The function [tex]\( g\left(\frac{1}{3} x\right) \)[/tex] does not change the range of [tex]\( g(x) \)[/tex], because only the input [tex]\( x \)[/tex] is being scaled, not the output. Thus, the range remains:
[tex]\[ -16 \leq y \leq 8 \][/tex]
(b) For the function [tex]\( y = 4g(x) \)[/tex]:
Domain:
The domain of [tex]\( y = 4g(x) \)[/tex] is the same as the original domain of [tex]\( g(x) \)[/tex], since the transformation involves only scaling the output, not the input. Therefore, the domain is:
[tex]\[ -2 \leq x \leq 11 \][/tex]
Range:
Since we are multiplying the output of [tex]\( g(x) \)[/tex] by 4, we need to scale the original range:
[tex]\[ y = 4 \cdot (-16) \quad \text{to} \quad y = 4 \cdot 8 \][/tex]
[tex]\[ -64 \leq y \leq 32 \][/tex]
So, the range is:
[tex]\[ -64 \leq y \leq 32 \][/tex]
Summarizing, the solutions for the domains and ranges are:
(a) For [tex]\( y = g\left(\frac{1}{3} x\right) \)[/tex]:
- Domain: [tex]\( -6 \leq x \leq 33 \)[/tex]
- Range: [tex]\( -16 \leq y \leq 8 \)[/tex]
(b) For [tex]\( y = 4g(x) \)[/tex]:
- Domain: [tex]\( -2 \leq x \leq 11 \)[/tex]
- Range: [tex]\( -64 \leq y \leq 32 \)[/tex]