To determine the domain and range of the function [tex]\( g(x) = -\frac{1}{3} f(x + 5) + 5 \)[/tex] given the domain and range of [tex]\( f(x) \)[/tex], you'll need to follow these steps:
1. Determine the domain of [tex]\( g(x) \)[/tex]:
- Original domain of [tex]\( f(x) \)[/tex]: [tex]\( [-3, 18] \)[/tex]
- The function [tex]\( g(x) \)[/tex] is derived from [tex]\( f(x) \)[/tex] by applying a horizontal shift of [tex]\( -5 \)[/tex]. This means that the input values for [tex]\( x \)[/tex] in [tex]\( f(x + 5) \)[/tex] correspond to [tex]\( x + 5 \)[/tex] values in the original domain of [tex]\( f(x) \)[/tex]. Therefore:
[tex]\[
\text{New domain} = [(-3) - 5, 18 - 5] = [-8, 13]
\][/tex]
2. Determine the range of [tex]\( g(x) \)[/tex]:
- Original range of [tex]\( f(x) \)[/tex]: [tex]\( [-12, 9] \)[/tex]
- The transformation applied to [tex]\( f(x) \)[/tex] involves two steps:
- Scaling by [tex]\( -\frac{1}{3} \)[/tex]: Since we are scaling the function by [tex]\( -\frac{1}{3} \)[/tex], we need to multiply the endpoints of the range of [tex]\( f(x) \)[/tex] by [tex]\( -\frac{1}{3} \)[/tex]. This changes the range to:
[tex]\[
\left[-\frac{1}{3} \times 9, -\frac{1}{3} \times (-12)\right] = [-3, 4]
\][/tex]
- Shifting by [tex]\( +5 \)[/tex]: After scaling, we add [tex]\( 5 \)[/tex] to each value in the range to find the new range:
[tex]\[
[-3 + 5, 4 + 5] = [2, 9]
\][/tex]
Therefore, the domain and range of the function [tex]\( g(x) \)[/tex] are as follows:
- Domain: [tex]\( [-8, 13] \)[/tex]
- Range: [tex]\( [2, 9] \)[/tex]
The final answer in interval notation is:
- Domain: [tex]\( [-8, 13] \)[/tex]
- Range: [tex]\( [2, 9] \)[/tex]
