Use the given function $$ f $$ to answer parts (a) through (f) below. $$ f(x)=e^{x+2}+1 $$

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**(a) Domain**  
The function is  
$$
f(x) = e^{x+2} + 1.
$$  
Since the exponential function $$ e^{x+2} $$ is defined for every real number, the domain of $$ f $$ is  
$$
\{ x \in \mathbb{R} \}.
$$

**(b) Range**  
Because $$ e^{x+2} $$ is always positive, we have  
$$
e^{x+2} > 0 \quad \text{for all } x.
$$  
Thus,  
$$
f(x) = e^{x+2} + 1 > 0 + 1 = 1.
$$  
Moreover, as $$ x \to -\infty $$, $$ e^{x+2} \to 0 $$ so that $$ f(x) \to 1 $$ (but never equals 1), and as $$ x \to \infty $$, $$ e^{x+2} \to \infty $$ so that $$ f(x) \to \infty $$. Therefore, the range is  
$$
(1, \infty).
$$

**(c) $$ x $$-intercepts**  
To find the $$ x $$-intercepts, set $$ f(x) = 0 $$:  
$$
e^{x+2} + 1 = 0.
$$  
Subtract 1 from both sides:  
$$
e^{x+2} = -1.
$$  
Since $$ e^{x+2} $$ is always positive, there is no solution to this equation. Hence, the function has no $$ x $$-intercepts.

**(d) $$ y $$-intercept**  
The $$ y $$-intercept occurs when $$ x = 0 $$. Substitute $$ x = 0 $$ into $$ f(x) $$:  
$$
f(0) = e^{0+2} + 1 = e^2 + 1.
$$  
Thus, the $$ y $$-intercept is at  
$$
(0, e^2+1).
$$

**(e) Inverse Function**  
To find the inverse, start with  
$$
y = e^{x+2} + 1.
$$  
1. Subtract 1 from both sides:  
   $$
   y - 1 = e^{x+2}.
   $$
2. Take the natural logarithm of both sides:  
   $$
   \ln(y-1) = x + 2.
   $$
3. Solve for $$ x $$:  
   $$
   x = \ln(y-1) - 2.
   $$
Now, replace $$ y $$ with $$ x $$ for the inverse function $$ f^{-1}(x) $$:  
$$
f^{-1}(x) = \ln(x-1) - 2.
$$
Note: The domain of $$ f^{-1} $$ is the range of $$ f $$, which is $$ (1, \infty) $$.

**(f) Transformations of $$ f(x) $$**  
The function can be written as  
$$
f(x) = e^{x+2} + 1.
$$  
This represents two transformations of the basic exponential function $$ e^x $$:  
1. A horizontal shift: The exponent $$ x+2 $$ indicates a shift $$ 2 $$ units to the left.  
2. A vertical shift: The $$ +1 $$ outside the exponential moves the graph up by $$ 1 $$ unit.

Thus, $$ f(x) $$ is the graph of $$ e^x $$ shifted $$ 2 $$ units left and $$ 1 $$ unit up.
**(a) Domain:** All real numbers **(b) Range:** $$ (1, \infty) $$ **(c) $$ x $$-intercepts:** None **(d) $$ y $$-intercept:** $$ (0, e^2 + 1) $$ **(e) Inverse Function:** $$ f^{-1}(x) = \ln(x - 1) - 2 $$ **(f) Transformations:** Shifted 2 units left and 1 unit up from $$ e^x $$.

Use the given function \( f \) to answer parts (a) through (f) below. \[ f(x)=e^{x+2}+1 \]

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