Define the function [tex]\( f(x) \)[/tex] as follows:

[tex]\[ f(x) = \begin{cases}
x - 2 & \text{if } -\infty \ \textless \ x \leq 0 \\
3^x - 1 & \text{if } 0 \ \textless \ x \ \textless \ \infty
\end{cases} \][/tex]



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].