Question

look at the table of values for the functions $f(x) = 2x + 1$ and $g(x) = 3^x - 3$.

$x$	$f(x)$	$g(x)$
$0$	$1$	$-2$
$1$	$3$	$0$
$2$	$5$	$6$
$3$	$7$	$24$

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

Explanation:

Step1: Analyze \( x = 0 \) and \( x = 1 \)

At \( x = 0 \), \( f(0)=1 \), \( g(0)= - 2 \), so \( f(0)>g(0) \). At \( x = 1 \), \( f(1)=3 \), \( g(1)=0 \), so \( f(1)>g(1) \)? Wait, no, wait, \( g(1)=3^{1}-3 = 0 \), \( f(1)=2\times1 + 1=3 \). Wait, no, wait, let's check \( x = 1 \) and \( x = 2 \). At \( x = 1 \), \( f(1)=3 \), \( g(1)=0 \) (so \( f(1)>g(1) \)). At \( x = 2 \), \( f(2)=5 \), \( g(2)=3^{2}-3=6 \) (so \( f(2)<g(2) \)). So between \( x = 1 \) and \( x = 2 \), the function \( f(x) \) goes from 3 to 5, and \( g(x) \) goes from 0 to 6. Since \( f(x) \) is linear and \( g(x) \) is exponential, and at \( x = 1 \), \( f(x)>g(x) \), at \( x = 2 \), \( f(x)<g(x) \), by Intermediate Value Theorem, there is a solution between \( x = 1 \) and \( x = 2 \). Wait, but let's re - check the table. Wait, at \( x = 1 \), \( f(1)=3 \), \( g(1)=0 \) (so \( f(1)>g(1) \)). At \( x = 2 \), \( f(2)=5 \), \( g(2)=6 \) (so \( f(2)<g(2) \)). So the equation \( f(x)=g(x) \) will have a solution between \( x = 1 \) and \( x = 2 \) because the two functions cross each other in that interval (one is above, then below).

Step2: Eliminate other options

• For \( x = 0 \): \( f(0)=1 \), \( g(0)=-2 \), not equal.
• For between \( x = 0 \) and \( x = 1 \): At \( x = 0 \), \( f(0)=1 \), \( g(0)=-2 \); at \( x = 1 \), \( f(1)=3 \), \( g(1)=0 \). Both times \( f(x)>g(x) \), so no crossing here.
• For \( x = 1 \): \( f(1)=3 \), \( g(1)=0 \), not equal.

Answer:

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