The given function is $f(x) = e^x + \ln(x + 3)$, and we need to determine the range of this function for $c \in [-2.5, 2]$.

Steps to solve:

1. Domain of the function:

   • The function involves $\ln(x + 3)$, which requires $x + 3 > 0$, so $x > -3$.
   • Given $-2.5 \leq c \leq 2$, $x + 3 \geq 0.5$. Hence, the domain for $c$ is valid within this range.

2. Behavior of each component:

   • $e^x$ is an increasing exponential function, and for $x \in [-2.5, 2]$, it ranges from $e^{-2.5}$ to $e^2$.
   • $\ln(x + 3)$ is a logarithmic function that increases as $x + 3$ increases. For $x \in [-2.5, 2]$, $x + 3 \in [0.5, 5]$, so $\ln(x + 3)$ ranges from $\ln(0.5)$ to $\ln(5)$.

3. Finding the range:

   • At $x = -2.5$: $f(-2.5) = e^{-2.5} + \ln(0.5)$.
   • At $x = 2$: $f(2) = e^2 + \ln(5)$.

4. Approximations:

   • $e^{-2.5} \approx 0.082$, $\ln(0.5) \approx -0.693$. So, $f(-2.5) \approx 0.082 - 0.693 = -0.611$.
   • $e^2 \approx 7.389$, $\ln(5) \approx 1.609$. So, $f(2) \approx 7.389 + 1.609 = 8.998$.

Thus, the range of $f(x)$ for $c \in [-2.5, 2]$ is approximately $(-0.611, 8.998)$.

The final answer in exact symbolic form is:

$f(c) \in (-0.611, 8.998)$

Would you like more details or have any questions?

Related questions:

1. How does the range of a logarithmic function vary with its argument?
2. What is the behavior of the exponential function at negative values of $x$?
3. Can we generalize the range of this function for different intervals of $x$?
4. What impact does shifting the logarithmic function by 3 have on its range?
5. How would you approach finding the range of a composite function like $e^x + \ln(x + 3)$?

Tip:

When finding the range of functions that combine exponentials and logarithms, always consider how both components behave over the interval you are interested in.