Answer :
Certainly! Let's explore the function [tex]\( f(x) \)[/tex]:
The function [tex]\( f(x) \)[/tex] is defined piecewise over two intervals. Let's write it out clearly:
[tex]\[ f(x) = \begin{cases} x - 2 & \text{if } x \leq 0 \\ 3^x - 1 & \text{if } x > 0 \end{cases} \][/tex]
### Analyzing the two pieces of the function:
#### For [tex]\( x \leq 0 \)[/tex]:
In this interval, [tex]\( f(x) = x - 2 \)[/tex].
- This is a linear function with a slope of 1 and a y-intercept at -2.
- When [tex]\( x = 0 \)[/tex], the value of [tex]\( f(x) = 0 - 2 = -2 \)[/tex].
#### For [tex]\( x > 0 \)[/tex]:
In this interval, [tex]\( f(x) = 3^x - 1 \)[/tex].
- This is an exponential function with a base of 3, shifted downward by 1.
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] grows exponentially.
- When [tex]\( x = 0 \)[/tex], hypothetically [tex]\( 3^x - 1 = 3^0 - 1 = 1 - 1 = 0 \)[/tex]. However, through our definition, this result is not used because the point where [tex]\( x=0 \)[/tex] falls into the other piece of the function.
### Continuity of the function at [tex]\( x = 0 \)[/tex]:
We need to determine if the function is continuous at [tex]\( x = 0 \)[/tex].
- From the left (as [tex]\( x \)[/tex] approaches 0 from the negative side), we use [tex]\( f(x) = x - 2 \)[/tex]. Thus, [tex]\(\lim_{{x \to 0^{-}}} f(x) = 0 - 2 = -2\)[/tex].
- From the right (as [tex]\( x \)[/tex] approaches 0 from the positive side), we use [tex]\( f(x) = 3^x - 1 \)[/tex]. Thus, [tex]\(\lim_{{x \to 0^{+}}} f(x) = 3^0 - 1 = 1 - 1 = 0\)[/tex].
Here, observe there is a discontinuity at [tex]\( x = 0 \)[/tex] since:
[tex]\[ \lim_{{x \to 0^{-}}} f(x) \neq \lim_{{x \to 0^{+}}} f(x). \][/tex]
### Evaluating specific points:
Let's evaluate a few points to understand the behavior of the function better:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -1 - 2 = -3 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 3^1 - 1 = 3 - 1 = 2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 3^2 - 1 = 9 - 1 = 8 \][/tex]
### Graphical Representation:
1. On [tex]\( x \leq 0 \)[/tex] side, the graph is a straight line with a negative y-intercept.
2. As [tex]\( x \)[/tex] becomes positive, the graph shifts to the exponential behavior.
From these evaluations:
- When [tex]\( x = 0 \)[/tex], the value is taken from [tex]\( x - 2 \)[/tex] part, giving [tex]\( f(0) = -2 \)[/tex].
In summary, the function [tex]\( f(x) \)[/tex] as defined showcases a piecewise behavior transitioning from a linear function for [tex]\( x \leq 0 \)[/tex] to an exponential one for [tex]\( x > 0 \)[/tex], with a clear discontinuity at [tex]\( x = 0 \)[/tex].
The function [tex]\( f(x) \)[/tex] is defined piecewise over two intervals. Let's write it out clearly:
[tex]\[ f(x) = \begin{cases} x - 2 & \text{if } x \leq 0 \\ 3^x - 1 & \text{if } x > 0 \end{cases} \][/tex]
### Analyzing the two pieces of the function:
#### For [tex]\( x \leq 0 \)[/tex]:
In this interval, [tex]\( f(x) = x - 2 \)[/tex].
- This is a linear function with a slope of 1 and a y-intercept at -2.
- When [tex]\( x = 0 \)[/tex], the value of [tex]\( f(x) = 0 - 2 = -2 \)[/tex].
#### For [tex]\( x > 0 \)[/tex]:
In this interval, [tex]\( f(x) = 3^x - 1 \)[/tex].
- This is an exponential function with a base of 3, shifted downward by 1.
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] grows exponentially.
- When [tex]\( x = 0 \)[/tex], hypothetically [tex]\( 3^x - 1 = 3^0 - 1 = 1 - 1 = 0 \)[/tex]. However, through our definition, this result is not used because the point where [tex]\( x=0 \)[/tex] falls into the other piece of the function.
### Continuity of the function at [tex]\( x = 0 \)[/tex]:
We need to determine if the function is continuous at [tex]\( x = 0 \)[/tex].
- From the left (as [tex]\( x \)[/tex] approaches 0 from the negative side), we use [tex]\( f(x) = x - 2 \)[/tex]. Thus, [tex]\(\lim_{{x \to 0^{-}}} f(x) = 0 - 2 = -2\)[/tex].
- From the right (as [tex]\( x \)[/tex] approaches 0 from the positive side), we use [tex]\( f(x) = 3^x - 1 \)[/tex]. Thus, [tex]\(\lim_{{x \to 0^{+}}} f(x) = 3^0 - 1 = 1 - 1 = 0\)[/tex].
Here, observe there is a discontinuity at [tex]\( x = 0 \)[/tex] since:
[tex]\[ \lim_{{x \to 0^{-}}} f(x) \neq \lim_{{x \to 0^{+}}} f(x). \][/tex]
### Evaluating specific points:
Let's evaluate a few points to understand the behavior of the function better:
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -1 - 2 = -3 \][/tex]
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 3^1 - 1 = 3 - 1 = 2 \][/tex]
- For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 3^2 - 1 = 9 - 1 = 8 \][/tex]
### Graphical Representation:
1. On [tex]\( x \leq 0 \)[/tex] side, the graph is a straight line with a negative y-intercept.
2. As [tex]\( x \)[/tex] becomes positive, the graph shifts to the exponential behavior.
From these evaluations:
- When [tex]\( x = 0 \)[/tex], the value is taken from [tex]\( x - 2 \)[/tex] part, giving [tex]\( f(0) = -2 \)[/tex].
In summary, the function [tex]\( f(x) \)[/tex] as defined showcases a piecewise behavior transitioning from a linear function for [tex]\( x \leq 0 \)[/tex] to an exponential one for [tex]\( x > 0 \)[/tex], with a clear discontinuity at [tex]\( x = 0 \)[/tex].
