The function provided is $f(x) = 3^x + x$, and we are asked to find the values for $f(-2)$, $f(0)$, and $f(3)$.

Step-by-step solutions:

1. For $f(-2)$: $f(-2) = 3^{-2} + (-2)$ $3^{-2} = \frac{1}{3^2} = \frac{1}{9}$ So, $f(-2) = \frac{1}{9} - 2 = \frac{1}{9} - \frac{18}{9} = \frac{-17}{9}$ Therefore, $f(-2) = \frac{-17}{9}$.

2. For $f(0)$: $f(0) = 3^0 + 0 = 1 + 0 = 1$ Therefore, $f(0) = 1$.

3. For $f(3)$: $f(3) = 3^3 + 3 = 27 + 3 = 30$ Therefore, $f(3) = 30$.

Summary of the values:

• $f(-2) = \frac{-17}{9}$
• $f(0) = 1$
• $f(3) = 30$

You can now place these values into the corresponding boxes.

Do you need more details on the calculations or any further explanation?

5 Related Questions:

1. How do you evaluate exponential functions for negative exponents?
2. What are the properties of exponents that help simplify expressions like $3^{-2}$?
3. How would the function change if the exponent had a variable like $x^2$?
4. Can the function $f(x) = 3^x + x$ ever be undefined? Why or why not?
5. How do we determine the asymptotic behavior of exponential functions as $x$ approaches infinity?

Tip:

Always check whether the base of an exponent is positive or negative, as it can change the overall result when raising to negative powers.