Answer :

Let's analyze the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] to determine its domain and range.

### Domain

The domain of a function is the set of all possible input values (x-values) that the function can accept. For the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex]:

1. The expression [tex]\( 5^{x-3} \)[/tex] is an exponential function. Exponential functions [tex]\( a^x \)[/tex] (with the base [tex]\( a > 0 \)[/tex]) are defined for all real numbers [tex]\( x \)[/tex].

Therefore, the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] is defined for all real numbers [tex]\( x \)[/tex].

So, the domain of the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
which means all real numbers.

### Range

The range of a function is the set of all possible output values (y-values) that the function can produce. For the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex]:

1. Consider the term [tex]\( 5^{x-3} \)[/tex]. The expression [tex]\( 5^{x-3} \)[/tex] represents an exponential function where the base is positive ([tex]\( 5 > 1 \)[/tex]). Exponential functions of this form are always positive: [tex]\( 5^{x-3} > 0 \)[/tex] for all real numbers [tex]\( x \)[/tex].

2. Since [tex]\( 5^{x-3} \)[/tex] is always positive, adding 1 to it will make the expression [tex]\( 5^{x-3} + 1 \)[/tex] always greater than 1. Specifically:
[tex]\[ f(x) = 5^{x-3} + 1 > 1 \][/tex]
for all real numbers [tex]\( x \)[/tex].

Therefore, [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] can take any value greater than 1.

So, the range of the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] is:
[tex]\[ (1, \infty) \][/tex]
which means all real numbers greater than 1.

### Summary

- The domain of the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
- The range of the function [tex]\( f(x) = 5^{x-3} + 1 \)[/tex] is:
[tex]\[ (1, \infty) \][/tex]