Question

look at the table of values for the functions $f(x) = |x| - 2$ and $g(x) = \left(\frac{1}{3}\
ight)^x$.

$x$	$f(x)$	$g(x)$
$0$	$-2$	$1$
$1$	$-1$	$\frac{1}{3}$
$2$	$0$	$\frac{1}{9}$
$3$	$1$	$\frac{1}{27}$

based on the values in the table, where does the equation $f(x) = g(x)$ have a solution?
$x = 1$ $x = 2$ between $x = 1$ and $x = 2$ between $x = 2$ and $x = 3$

Explanation:

Step1: Analyze \( x = 1 \)

At \( x = 1 \), \( f(1)= - 1 \) and \( g(1)=\frac{1}{3}\). So \( f(1)<g(1) \) (since \(-1<\frac{1}{3}\)).

Step2: Analyze \( x = 2 \)

At \( x = 2 \), \( f(2) = 0 \) and \( g(2)=\frac{1}{9}\). So \( f(2)>g(2) \) (since \(0>\frac{1}{9}\)).

Step3: Apply Intermediate Value Theorem

Since \( f(x) \) and \( g(x) \) are continuous functions (absolute - value function and exponential function are continuous), and \( f(1)<g(1) \) while \( f(2)>g(2) \), by the Intermediate Value Theorem, there must be some \( c \) in the interval \((1,2)\) such that \( f(c)=g(c) \).

For \( x = 1 \): \( f(1)=-1
eq g(1)=\frac{1}{3} \). For \( x = 2 \): \( f(2) = 0
eq g(2)=\frac{1}{9} \). For the interval between \( x = 2 \) and \( x = 3 \): \( f(3)=1\) and \( g(3)=\frac{1}{27}\), \( f(3)>g(3) \) and at \( x = 2 \), \( f(2)>g(2) \), so no crossing here.

Answer:

between \( x = 1 \) and \( x = 2 \)