From the table provided in the image, we are tasked to determine where $f(x) = g(x)$ has a solution, given:

• $f(x) = |x| + 5$
• $g(x) = \left(\frac{1}{2}\right)^x$

Table of values:

Step-by-Step Solution:

1. Where $f(x) = g(x)$:

   • At $x = -3$: $f(x) = 8$ and $g(x) = 8$, so they are equal at $x = -3$.

2. Between intervals:

   • For $x = -2$: $f(x) = 7$ and $g(x) = 4$. Since $f(x) > g(x)$, there is no equality here.
   • For $x = -1$: $f(x) = 6$ and $g(x) = 2$. Again, $f(x) > g(x)$.
   • For $x = 0$: $f(x) = 5$ and $g(x) = 1$. Still, $f(x) > g(x)$.

3. Conclusion:

   • $f(x) = g(x)$ at $x = -3$.
   • The correct answer is $x = -3$.

Do you need further clarification or a deeper breakdown of the calculations?

Related Questions:

1. What happens to $f(x) = |x| + 5$ as $x \to \infty$?
2. What happens to $g(x) = \left(\frac{1}{2}\right)^x$ as $x \to -\infty$?
3. How can you solve $f(x) = g(x)$ algebraically instead of using a table?
4. How does $g(x)$ compare to $f(x)$ for $x > 1$?
5. Can $f(x)$ and $g(x)$ intersect more than once for $x \in \mathbb{R}$?

Tip:

Always analyze the behavior of the functions at extreme values (like $x \to \infty$ or $x \to -\infty$) to understand their trends!