Look at the table of values for the functions f(x) = x^2 - 2 and g(x) = |x - 2|. Ba...

1. Understand the Goal: The problem asks us to find where the equation $f(x) = g(x)$ has a solution, based on the provided table of values for the functions $f(x) = x^2 - 2$ and $g(x) = |x - 2|$. A solution exists when the values of $f(x)$ and $g(x)$ are equal for the same value of $x$.

1. Examine the Table: We need to look for a row in the table where the value in the $f(x)$ column is equal to the value in the $g(x)$ column.

• For $x = 0$: $f(0) = -2$ and $g(0) = 2$. Here, $f(0) \neq g(0)$.
• For $x = 1$: $f(1) = -1$ and $g(1) = 1$. Here, $f(1) \neq g(1)$.
• For $x = 2$: $f(2) = 2$ and $g(2) = 0$. Here, $f(2) \neq g(2)$.
• For $x = 3$: $f(3) = 7$ and $g(3) = 1$. Here, $f(3) \neq g(3)$.
• For $x = 4$: $f(4) = 14$ and $g(4) = 2$. Here, $f(4) \neq g(4)$.

1. Analyze Function Behavior Between Points: Since we don't see an exact match in the table, we need to check if a solution might exist between the given integer values of $x$. We are looking for an interval where $f(x)$ and $g(x)$ cross each other, meaning one function's value is greater than the other at one endpoint of the interval and less than the other at the other endpoint.

- Interval between $x=0$ and $x=1$:
At $x=0$, $f(0) = -2$ and $g(0) = 2$. So, $g(0) > f(0)$.
At $x=1$, $f(1) = -1$ and $g(1) = 1$. So, $g(1) > f(1)$.
Since $g(x)$ is greater than $f(x)$ at both ends of this interval, and both functions are continuous, it's unlikely they intersect within this interval unless they touch and separate without crossing, which is not the case here as $g(x)$ is consistently above $f(x)$.

- Interval between $x=1$ and $x=2$:
At $x=1$, $f(1) = -1$ and $g(1) = 1$. So, $g(1) > f(1)$.
At $x=2$, $f(2) = 2$ and $g(2) = 0$. So, $f(2) > g(2)$.
In this interval, $g(x)$ starts above $f(x)$ (at $x=1$) and ends up below $f(x)$ (at $x=2$). Because both functions are continuous, the Intermediate Value Theorem guarantees that there must be at least one point $x$ between 1 and 2 where $f(x) = g(x)$.

- Interval between $x=2$ and $x=3$:
At $x=2$, $f(2) = 2$ and $g(2) = 0$. So, $f(2) > g(2)$.
At $x=3$, $f(3) = 7$ and $g(3) = 1$. So, $f(3) > g(3)$.
Here, $f(x)$ is greater than $g(x)$ at both ends of the interval, so no solution is guaranteed in this interval.

- Interval between $x=3$ and $x=4$:
At $x=3$, $f(3) = 7$ and $g(3) = 1$. So, $f(3) > g(3)$.
At $x=4$, $f(4) = 14$ and $g(4) = 2$. So, $f(4) > g(4)$.
Here, $f(x)$ is greater than $g(x)$ at both ends of the interval, so no solution is guaranteed in this interval.

1. Conclusion: Based on the analysis of the intervals, the equation $f(x) = g(x)$ has a solution between $x = 1$ and $x = 2$.