Answer :
Let's go through the given piecewise function [tex]\( f(x) \)[/tex] and solve each part step-by-step:
The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = \begin{cases} 3^{x-1} - 4 & \text{for } x \leq 3 \\ \frac{15}{x} & \text{for } x > 3 \end{cases} \][/tex]
### Part A: Range of [tex]\( f(x) \)[/tex]
To find the range of [tex]\( f(x) \)[/tex], we need to analyze both parts of the piecewise function separately:
1. For [tex]\( x \leq 3 \)[/tex]:
The function is [tex]\( f(x) = 3^{x-1} - 4 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} \)[/tex] approaches [tex]\( 0 \)[/tex] from above since [tex]\( 3^{-\infty} = 0 \)[/tex].
- Therefore, [tex]\( 3^{x-1} - 4 \)[/tex] will approach [tex]\( -4 \)[/tex] from above.
- When [tex]\( x = 3 \)[/tex], [tex]\( f(3) = 3^{3-1} - 4 = 3^2 - 4 = 9 - 4 = 5 \)[/tex].
- As [tex]\( x \)[/tex] increases towards [tex]\( 3 \)[/tex], [tex]\( 3^{x-1} - 4 \)[/tex] increases exponentially.
Hence, the output of [tex]\( f(x) \)[/tex] for [tex]\( x \leq 3 \)[/tex] can vary from [tex]\( -\infty \)[/tex] (as [tex]\( x \to -\infty \)[/tex]) to [tex]\( 5 \)[/tex] (at [tex]\( x = 3 \)[/tex]).
2. For [tex]\( x > 3 \)[/tex]:
The function is [tex]\( f(x) = \frac{15}{x} \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \)[/tex] approaches [tex]\( 0 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 3 \)[/tex] from the right, [tex]\( \frac{15}{x} \)[/tex] approaches [tex]\( \frac{15}{3} = 5 \)[/tex].
Therefore, the output of [tex]\( f(x) \)[/tex] for [tex]\( x > 3 \)[/tex] can vary from [tex]\( 0 \)[/tex] to [tex]\( 5 \)[/tex].
Combining both sections of the piecewise function, we can see that the function [tex]\( f(x) \)[/tex] takes any value from [tex]\( -\infty \)[/tex] to [tex]\( 5 \)[/tex] (inclusive of [tex]\( 0 \)[/tex] but approaching [tex]\( 0 \)[/tex] as [tex]\( x\to\infty\)[/tex]).
Hence, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
### Part B: Asymptotes of [tex]\( f(x) \)[/tex]
1. Vertical Asymptote:
- For [tex]\( x \leq 3 \)[/tex], the expression [tex]\( 3^{x-1} - 4 \)[/tex] does not have any vertical asymptotes.
- For [tex]\( x > 3 \)[/tex], the expression [tex]\( \frac{15}{x} \)[/tex] would have a vertical asymptote at [tex]\( x = 0 \)[/tex]; however [tex]\( x = 0 \)[/tex] is outside the domain of this part of the function ([tex]\( x > 3 \)[/tex]). Therefore, within the given domain, there are no vertical asymptotes.
2. Horizontal Asymptote:
- For [tex]\( x \leq 3 \)[/tex], as [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} - 4 \to -4 \)[/tex].
- For [tex]\( x > 3 \)[/tex], as [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \to 0 \)[/tex].
Hence, the horizontal asymptote is [tex]\( y = 0 \)[/tex] as [tex]\( x \to \infty \)[/tex].
Hence, there is no vertical asymptote and there is a horizontal asymptote at:
[tex]\[ y = 0 \][/tex]
### Part C: End Behavior of [tex]\( f(x) \)[/tex]
The end behavior of the function describes how [tex]\( f(x) \)[/tex] behaves as [tex]\( x \)[/tex] approaches [tex]\( \pm \infty \)[/tex].
1. As [tex]\( x \to -\infty \)[/tex] (for [tex]\( x \leq 3 \)[/tex]):
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} \to 0 \)[/tex] and hence [tex]\( 3^{x-1} - 4 \to -4 \)[/tex].
Therefore, as [tex]\( x \to -\infty \)[/tex]:
[tex]\[ f(x) \to -4 \][/tex]
2. As [tex]\( x \to \infty \)[/tex] (for [tex]\( x > 3 \)[/tex]):
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \to 0 \)[/tex].
Therefore, as [tex]\( x \to \infty \)[/tex]:
[tex]\[ f(x) \to 0 \][/tex]
Summarizing the end behavior:
- As [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to \infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
### Final Solution Recap:
- Range of [tex]\( f(x) \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Vertical Asymptote: None
- Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
- End Behavior:
- As [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to \infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
The function [tex]\( f(x) \)[/tex] is defined as:
[tex]\[ f(x) = \begin{cases} 3^{x-1} - 4 & \text{for } x \leq 3 \\ \frac{15}{x} & \text{for } x > 3 \end{cases} \][/tex]
### Part A: Range of [tex]\( f(x) \)[/tex]
To find the range of [tex]\( f(x) \)[/tex], we need to analyze both parts of the piecewise function separately:
1. For [tex]\( x \leq 3 \)[/tex]:
The function is [tex]\( f(x) = 3^{x-1} - 4 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} \)[/tex] approaches [tex]\( 0 \)[/tex] from above since [tex]\( 3^{-\infty} = 0 \)[/tex].
- Therefore, [tex]\( 3^{x-1} - 4 \)[/tex] will approach [tex]\( -4 \)[/tex] from above.
- When [tex]\( x = 3 \)[/tex], [tex]\( f(3) = 3^{3-1} - 4 = 3^2 - 4 = 9 - 4 = 5 \)[/tex].
- As [tex]\( x \)[/tex] increases towards [tex]\( 3 \)[/tex], [tex]\( 3^{x-1} - 4 \)[/tex] increases exponentially.
Hence, the output of [tex]\( f(x) \)[/tex] for [tex]\( x \leq 3 \)[/tex] can vary from [tex]\( -\infty \)[/tex] (as [tex]\( x \to -\infty \)[/tex]) to [tex]\( 5 \)[/tex] (at [tex]\( x = 3 \)[/tex]).
2. For [tex]\( x > 3 \)[/tex]:
The function is [tex]\( f(x) = \frac{15}{x} \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \)[/tex] approaches [tex]\( 0 \)[/tex].
- As [tex]\( x \)[/tex] approaches [tex]\( 3 \)[/tex] from the right, [tex]\( \frac{15}{x} \)[/tex] approaches [tex]\( \frac{15}{3} = 5 \)[/tex].
Therefore, the output of [tex]\( f(x) \)[/tex] for [tex]\( x > 3 \)[/tex] can vary from [tex]\( 0 \)[/tex] to [tex]\( 5 \)[/tex].
Combining both sections of the piecewise function, we can see that the function [tex]\( f(x) \)[/tex] takes any value from [tex]\( -\infty \)[/tex] to [tex]\( 5 \)[/tex] (inclusive of [tex]\( 0 \)[/tex] but approaching [tex]\( 0 \)[/tex] as [tex]\( x\to\infty\)[/tex]).
Hence, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ (-\infty, \infty) \][/tex]
### Part B: Asymptotes of [tex]\( f(x) \)[/tex]
1. Vertical Asymptote:
- For [tex]\( x \leq 3 \)[/tex], the expression [tex]\( 3^{x-1} - 4 \)[/tex] does not have any vertical asymptotes.
- For [tex]\( x > 3 \)[/tex], the expression [tex]\( \frac{15}{x} \)[/tex] would have a vertical asymptote at [tex]\( x = 0 \)[/tex]; however [tex]\( x = 0 \)[/tex] is outside the domain of this part of the function ([tex]\( x > 3 \)[/tex]). Therefore, within the given domain, there are no vertical asymptotes.
2. Horizontal Asymptote:
- For [tex]\( x \leq 3 \)[/tex], as [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} - 4 \to -4 \)[/tex].
- For [tex]\( x > 3 \)[/tex], as [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \to 0 \)[/tex].
Hence, the horizontal asymptote is [tex]\( y = 0 \)[/tex] as [tex]\( x \to \infty \)[/tex].
Hence, there is no vertical asymptote and there is a horizontal asymptote at:
[tex]\[ y = 0 \][/tex]
### Part C: End Behavior of [tex]\( f(x) \)[/tex]
The end behavior of the function describes how [tex]\( f(x) \)[/tex] behaves as [tex]\( x \)[/tex] approaches [tex]\( \pm \infty \)[/tex].
1. As [tex]\( x \to -\infty \)[/tex] (for [tex]\( x \leq 3 \)[/tex]):
- As [tex]\( x \to -\infty \)[/tex], [tex]\( 3^{x-1} \to 0 \)[/tex] and hence [tex]\( 3^{x-1} - 4 \to -4 \)[/tex].
Therefore, as [tex]\( x \to -\infty \)[/tex]:
[tex]\[ f(x) \to -4 \][/tex]
2. As [tex]\( x \to \infty \)[/tex] (for [tex]\( x > 3 \)[/tex]):
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{15}{x} \to 0 \)[/tex].
Therefore, as [tex]\( x \to \infty \)[/tex]:
[tex]\[ f(x) \to 0 \][/tex]
Summarizing the end behavior:
- As [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to \infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
### Final Solution Recap:
- Range of [tex]\( f(x) \)[/tex]: [tex]\( (-\infty, \infty) \)[/tex]
- Vertical Asymptote: None
- Horizontal Asymptote: [tex]\( y = 0 \)[/tex]
- End Behavior:
- As [tex]\( x \to -\infty \)[/tex]: [tex]\( f(x) \to -\infty \)[/tex]
- As [tex]\( x \to \infty \)[/tex]: [tex]\( f(x) \to 0 \)[/tex]
