Question

look at the table of values for the functions $f(x) = -2x + 2$ and $g(x) = \left(\frac{1}{4}\
ight)^x$.\
\

$x$	$f(x)$	$g(x)$	\
---	---	---	\
$-2$	$6$	$16$	\
$-1$	$4$	$4$	\
$0$	$2$	$1$	\
$1$	$0$	$\frac{1}{4}$	\
$2$	$-2$	$\frac{1}{16}$	\

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

Explanation:

Step1: Check x = -2

For \( x = -2 \), \( f(-2) = 6 \) and \( g(-2) = 16 \). Since \( 6
eq16 \), \( x = -2 \) is not a solution.

Step2: Check x = -1

For \( x = -1 \), \( f(-1) = 4 \) and \( g(-1) = 4 \). Here, \( f(-1)=g(-1) \), so \( x = -1 \) is a solution.

Step3: Analyze other intervals

• Between \( x=-2 \) and \( x=-1 \): At \( x=-2 \), \( f(x)=6 \), \( g(x)=16 \) (\( f(x)<g(x) \)); at \( x=-1 \), \( f(x)=4 \), \( g(x)=4 \) (\( f(x)=g(x) \)). But we already found \( x=-1 \) is a solution, so this interval is not the answer.
• Between \( x=-1 \) and \( x=0 \): At \( x=-1 \), \( f(x)=g(x)=4 \); at \( x = 0 \), \( f(0)=2 \), \( g(0)=1 \) (\( f(x)>g(x) \)). But \( x=-1 \) is a solution, so this interval is not the answer.

Answer:

\( x = -1 \)