To determine whether the given piecewise function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 2 \)[/tex], we need to check three conditions:
1. The left-hand limit as [tex]\( x \)[/tex] approaches 2.
2. The right-hand limit as [tex]\( x \)[/tex] approaches 2.
3. The value of the function at [tex]\( x = 2 \)[/tex].
The function [tex]\( f(x) \)[/tex] is defined as follows:
[tex]\[
f(x) =
\begin{cases}
2x - 1 & \text{if } x \leq 2 \\
3^{x-2} & \text{if } x > 2
\end{cases}
\][/tex]
### Step 1: Calculate the Left-Hand Limit
To find the left-hand limit as [tex]\( x \)[/tex] approaches 2, we use the expression [tex]\( 2x - 1 \)[/tex]:
[tex]\[
\lim_{{x \to 2^-}} f(x) = 2(2) - 1 = 4 - 1 = 3
\][/tex]
### Step 2: Calculate the Right-Hand Limit
To find the right-hand limit as [tex]\( x \)[/tex] approaches 2, we use the expression [tex]\( 3^{x-2} \)[/tex]:
[tex]\[
\lim_{{x \to 2^+}} f(x) = 3^{2-2} = 3^0 = 1
\][/tex]
### Step 3: Evaluate [tex]\( f(x) \)[/tex] at [tex]\( x = 2 \)[/tex]
Given that [tex]\( x \leq 2 \)[/tex] for [tex]\( f(x) = 2x - 1 \)[/tex], we substitute [tex]\( x = 2 \)[/tex]:
[tex]\[
f(2) = 2(2) - 1 = 4 - 1 = 3
\][/tex]
### Step 4: Check for Continuity
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 2 \)[/tex], the left-hand limit, the right-hand limit, and the function value at [tex]\( x = 2 \)[/tex] must all be equal.
- The left-hand limit is [tex]\( 3 \)[/tex].
- The right-hand limit is [tex]\( 1 \)[/tex].
- The function value at [tex]\( x = 2 \)[/tex] is [tex]\( 3 \)[/tex].
Since the left-hand limit ([tex]\( 3 \)[/tex]) and the right-hand limit ([tex]\( 1 \)[/tex]) are not equal, the function [tex]\( f(x) \)[/tex] is not continuous at [tex]\( x = 2 \)[/tex].
Therefore, [tex]\( f(x) \)[/tex] is not a continuous function.
