State the range for the function $$ f(x)=\pi^{x}+2 $$ : $$ (2, \infty) $$ $$ (0, \infty) $$ $$ (-2, \infty) $$ $$ (-\infty, 2) $$

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To determine the range of the function $$ f(x) = \pi^{x} + 2 $$, let's analyze its behavior:

1. **Base Function $$ \pi^{x} $$:**
   - Since $$ \pi > 1 $$, $$ \pi^{x} $$ is an exponential function that is always **positive**.
   - As $$ x $$ increases, $$ \pi^{x} $$ grows **exponentially** towards $$ +\infty $$.
   - As $$ x $$ decreases (i.e., $$ x $$ becomes more negative), $$ \pi^{x} $$ approaches **0** but never actually reaches it.

2. **Translating $$ \pi^{x} $$ by 2:**
   - Adding 2 shifts the entire graph of $$ \pi^{x} $$ **upwards** by 2 units.
   - Therefore, $$ f(x) = \pi^{x} + 2 $$ will always be **greater than 2**.

3. **Conclusion on the Range:**
   - The smallest value $$ f(x) $$ can approach is just above 2 (but never actually 2).
   - There is no upper limit as $$ \pi^{x} $$ can grow without bound.
   - Hence, the range of $$ f(x) $$ is $$ (2, \infty) $$.

**Answer:** $$ (2, \infty) $$
The range of $$ f(x) = \pi^{x} + 2 $$ is $$ (2, \infty) $$.

State the range for the function \( f(x)=\pi^{x}+2 \) : \( (2, \infty) \) \( (0, \infty) \) \( (-2, \infty) \) \( (-\infty, 2) \)

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