Question

complete the table of values for the functions $f(x) = |x - 6|$ and $g(x) = 2^x + 3$.
write your answers as whole numbers, decimals, or simplified fractions or mixed numbers.

$x$	$f(x)$	$g(x)$
$0$
$1$
$2$
$3$

based on the values in the table, where does the equation $f(x) = g(x)$ have a solution?

$x = 1$ $\quad$ between $x = 1$ and $x = 2$

$x = 2$ $\quad$ between $x = 2$ and $x = 3$

Explanation:

Part 1: Completing the table for \( f(x) = |x - 6| \) and \( g(x) = 2^x + 3 \)

For \( f(x) = |x - 6| \):

• When \( x = -1 \):

\( f(-1) = |-1 - 6| = |-7| = 7 \)

• When \( x = 0 \):

\( f(0) = |0 - 6| = |-6| = 6 \)

• When \( x = 1 \):

\( f(1) = |1 - 6| = |-5| = 5 \)

• When \( x = 2 \):

\( f(2) = |2 - 6| = |-4| = 4 \)

• When \( x = 3 \):

\( f(3) = |3 - 6| = |-3| = 3 \)

For \( g(x) = 2^x + 3 \):

• When \( x = -1 \):

\( g(-1) = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5 \)

• When \( x = 0 \):

\( g(0) = 2^0 + 3 = 1 + 3 = 4 \)

• When \( x = 1 \):

\( g(1) = 2^1 + 3 = 2 + 3 = 5 \)

• When \( x = 2 \):

\( g(2) = 2^2 + 3 = 4 + 3 = 7 \)

• When \( x = 3 \):

\( g(3) = 2^3 + 3 = 8 + 3 = 11 \)

Filled Table:

\( x \)	\( f(x) \)	\( g(x) \)
\( 0 \)	\( 6 \)	\( 4 \)
\( 1 \)	\( 5 \)	\( 5 \)
\( 2 \)	\( 4 \)	\( 7 \)
\( 3 \)	\( 3 \)	\( 11 \)

Part 2: Solving \( f(x) = g(x) \)

From the table:

• At \( x = 1 \): \( f(1) = 5 \) and \( g(1) = 5 \), so \( f(1) = g(1) \). Wait, but let’s check again:

Wait, \( g(1) = 2^1 + 3 = 5 \), and \( f(1) = 5 \). So \( f(1) = g(1) \), meaning \( x = 1 \) is a solution? Wait, but let’s verify the table again.

Wait, when \( x = 1 \):
\( f(1) = |1 - 6| = 5 \)
\( g(1) = 2^1 + 3 = 5 \)

Thus, \( f(1) = g(1) \), so \( x = 1 \) is a solution.

Final Answers:

Table Completion:

\( x \)	\( f(x) \)	\( g(x) \)
\( 0 \)	\( 6 \)	\( 4 \)
\( 1 \)	\( 5 \)	\( 5 \)
\( 2 \)	\( 4 \)	\( 7 \)
\( 3 \)	\( 3 \)	\( 11 \)

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

\( x = 1 \)

Answer:

Part 1: Completing the table for \( f(x) = |x - 6| \) and \( g(x) = 2^x + 3 \)

For \( f(x) = |x - 6| \):

• When \( x = -1 \):

\( f(-1) = |-1 - 6| = |-7| = 7 \)

• When \( x = 0 \):

\( f(0) = |0 - 6| = |-6| = 6 \)

• When \( x = 1 \):

\( f(1) = |1 - 6| = |-5| = 5 \)

• When \( x = 2 \):

\( f(2) = |2 - 6| = |-4| = 4 \)

• When \( x = 3 \):

\( f(3) = |3 - 6| = |-3| = 3 \)

For \( g(x) = 2^x + 3 \):

• When \( x = -1 \):

\( g(-1) = 2^{-1} + 3 = \frac{1}{2} + 3 = 3.5 \)

• When \( x = 0 \):

\( g(0) = 2^0 + 3 = 1 + 3 = 4 \)

• When \( x = 1 \):

\( g(1) = 2^1 + 3 = 2 + 3 = 5 \)

• When \( x = 2 \):

\( g(2) = 2^2 + 3 = 4 + 3 = 7 \)

• When \( x = 3 \):

\( g(3) = 2^3 + 3 = 8 + 3 = 11 \)

Filled Table:

\( x \)	\( f(x) \)	\( g(x) \)
\( 0 \)	\( 6 \)	\( 4 \)
\( 1 \)	\( 5 \)	\( 5 \)
\( 2 \)	\( 4 \)	\( 7 \)
\( 3 \)	\( 3 \)	\( 11 \)

Part 2: Solving \( f(x) = g(x) \)

From the table:

• At \( x = 1 \): \( f(1) = 5 \) and \( g(1) = 5 \), so \( f(1) = g(1) \). Wait, but let’s check again:

Wait, \( g(1) = 2^1 + 3 = 5 \), and \( f(1) = 5 \). So \( f(1) = g(1) \), meaning \( x = 1 \) is a solution? Wait, but let’s verify the table again.

Wait, when \( x = 1 \):
\( f(1) = |1 - 6| = 5 \)
\( g(1) = 2^1 + 3 = 5 \)

Thus, \( f(1) = g(1) \), so \( x = 1 \) is a solution.

Final Answers:

Table Completion:

\( x \)	\( f(x) \)	\( g(x) \)
\( 0 \)	\( 6 \)	\( 4 \)
\( 1 \)	\( 5 \)	\( 5 \)
\( 2 \)	\( 4 \)	\( 7 \)
\( 3 \)	\( 3 \)	\( 11 \)

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

\( x = 1 \)