Question

complete the table of values for the functions $f(x) = x^2 - 2$ and $g(x) = |x - 2|$. write your answers as whole numbers, decimals, or simplified fractions or mixed numbers. \

$$\begin{tabular}{|c|c|c|} \\hline \\( x \\) & \\( f(x) \\) & \\( g(x) \\) \\\\ \\hline 0 & \\( \\square \\) & \\( \\square \\) \\\\ \\hline 1 & \\( \\square \\) & \\( \\square \\) \\\\ \\hline 2 & \\( \\square \\) & \\( \\square \\) \\\\ \\hline 3 & \\( \\square \\) & \\( \\square \\) \\\\ \\hline 4 & \\( \\square \\) & \\( \\square \\) \\\\ \\hline \\end{tabular}$$

based on the values in the table, where does the equation $f(x) = g(x)$ have a solution? \\( x = 0 \\) \\( x = 1 \\) between $x = 0$ and $x = 1$ between $x = 1$ and $x = 2$

Explanation:

Part 1: Complete the table for \( f(x) = x^2 - 2 \) and \( g(x) = |x - 2| \)

For \( x = 0 \):

• \( f(0) = 0^2 - 2 = -2 \)
• \( g(0) = |0 - 2| = 2 \)

For \( x = 1 \):

• \( f(1) = 1^2 - 2 = -1 \)
• \( g(1) = |1 - 2| = 1 \)

For \( x = 2 \):

• \( f(2) = 2^2 - 2 = 2 \)
• \( g(2) = |2 - 2| = 0 \)

For \( x = 3 \):

• \( f(3) = 3^2 - 2 = 7 \)
• \( g(3) = |3 - 2| = 1 \)

For \( x = 4 \):

• \( f(4) = 4^2 - 2 = 14 \)
• \( g(4) = |4 - 2| = 2 \)

The completed table is:

\( x \)	\( f(x) \)	\( g(x) \)
1	-1	1
2	2	0
3	7	1
4	14	2

Part 2: Find where \( f(x) = g(x) \)

We analyze the values of \( f(x) \) and \( g(x) \) at each \( x \):

• At \( x = 0 \): \( f(0) = -2 \), \( g(0) = 2 \) ( \( f(x) < g(x) \) )
• At \( x = 1 \): \( f(1) = -1 \), \( g(1) = 1 \) ( \( f(x) < g(x) \) )
• At \( x = 2 \): \( f(2) = 2 \), \( g(2) = 0 \) ( \( f(x) > g(x) \) )

Since \( f(x) \) is increasing (quadratic) and \( g(x) \) is decreasing then increasing (absolute value), we check the intervals:

• Between \( x = 1 \) and \( x = 2 \): At \( x = 1 \), \( f(x) = -1 \), \( g(x) = 1 \) ( \( f < g \) ). At \( x = 2 \), \( f(x) = 2 \), \( g(x) = 0 \) ( \( f > g \) ). By the Intermediate Value Theorem, \( f(x) = g(x) \) has a solution between \( x = 1 \) and \( x = 2 \).

Final Answers

Table:

\( x \)	\( f(x) \)	\( g(x) \)
1	-1	1
2	2	0
3	7	1
4	14	2

Solution to \( f(x) = g(x) \):

The equation \( f(x) = g(x) \) has a solution between \( x = 1 \) and \( x = 2 \).

Answer:

Part 1: Complete the table for \( f(x) = x^2 - 2 \) and \( g(x) = |x - 2| \)

For \( x = 0 \):

• \( f(0) = 0^2 - 2 = -2 \)
• \( g(0) = |0 - 2| = 2 \)

For \( x = 1 \):

• \( f(1) = 1^2 - 2 = -1 \)
• \( g(1) = |1 - 2| = 1 \)

For \( x = 2 \):

• \( f(2) = 2^2 - 2 = 2 \)
• \( g(2) = |2 - 2| = 0 \)

For \( x = 3 \):

• \( f(3) = 3^2 - 2 = 7 \)
• \( g(3) = |3 - 2| = 1 \)

For \( x = 4 \):

• \( f(4) = 4^2 - 2 = 14 \)
• \( g(4) = |4 - 2| = 2 \)

The completed table is:

\( x \)	\( f(x) \)	\( g(x) \)
1	-1	1
2	2	0
3	7	1
4	14	2

Part 2: Find where \( f(x) = g(x) \)

We analyze the values of \( f(x) \) and \( g(x) \) at each \( x \):

• At \( x = 0 \): \( f(0) = -2 \), \( g(0) = 2 \) ( \( f(x) < g(x) \) )
• At \( x = 1 \): \( f(1) = -1 \), \( g(1) = 1 \) ( \( f(x) < g(x) \) )
• At \( x = 2 \): \( f(2) = 2 \), \( g(2) = 0 \) ( \( f(x) > g(x) \) )

Since \( f(x) \) is increasing (quadratic) and \( g(x) \) is decreasing then increasing (absolute value), we check the intervals:

• Between \( x = 1 \) and \( x = 2 \): At \( x = 1 \), \( f(x) = -1 \), \( g(x) = 1 \) ( \( f < g \) ). At \( x = 2 \), \( f(x) = 2 \), \( g(x) = 0 \) ( \( f > g \) ). By the Intermediate Value Theorem, \( f(x) = g(x) \) has a solution between \( x = 1 \) and \( x = 2 \).

Final Answers

Table:

\( x \)	\( f(x) \)	\( g(x) \)
1	-1	1
2	2	0
3	7	1
4	14	2

Solution to \( f(x) = g(x) \):

The equation \( f(x) = g(x) \) has a solution between \( x = 1 \) and \( x = 2 \).