Question

what is the initial value of the exponential function represented by the table? options: \\(\frac{1}{2}\\), \\(\frac{1}{8}\\), 1, \\(\frac{1}{4}\\) (table with x values -2, -1, 0, 1, 2 and f(x) values \\(\frac{1}{8}\\), \\(\frac{1}{4}\\), \\(\frac{1}{2}\\), 1, 2)

Explanation:

Step1: Recall exponential function form

An exponential function is of the form \( f(x) = ab^x \), where \( a \) is the initial value (when \( x = 0 \)) and \( b \) is the base.

Step2: Find \( f(0) \) from the table

Looking at the table, when \( x = 0 \), \( f(0)=\frac{1}{2} \)? Wait, no, wait the table: let's check again. Wait the table has \( x = 0 \), \( f(x)=\frac{1}{2} \)? Wait no, the table: first column \( x \), second \( f(x) \). For \( x = 0 \), the value of \( f(x) \) is \( \frac{1}{2} \)? Wait no, let's re - examine the table. Wait the user's table: when \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, the original table (from the image description): \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, the user's table: \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, the first row: \( x = - 2 \), \( f(x)=\frac{1}{8} \)? Wait no, the user's table: \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, let's look again. The table:

\( x=-2 \), \( f(x)=\frac{1}{8} \)

\( x=-1 \), \( f(x)=\frac{1}{4} \)

\( x = 0 \), \( f(x)=\frac{1}{2} \)

\( x = 1 \), \( f(x)=1 \)

\( x = 2 \), \( f(x)=2 \)

Wait, no, the initial value of an exponential function \( f(x)=ab^x \) is the value when \( x = 0 \), because when \( x = 0 \), \( f(0)=ab^0=a\times1=a \). So we need to find \( f(0) \) from the table. Looking at the table, when \( x = 0 \), \( f(0)=\frac{1}{2} \)? Wait no, the table (as per the image description) has \( x = 0 \), \( f(x)=\frac{1}{2} \)? Wait no, the user's table: let's check the rows. Wait the first row: \( x=-2 \), \( f(x)=\frac{1}{8} \); \( x=-1 \), \( f(x)=\frac{1}{4} \); \( x = 0 \), \( f(x)=\frac{1}{2} \); \( x = 1 \), \( f(x)=1 \); \( x = 2 \), \( f(x)=2 \). Wait, but the options are \( \frac{1}{2} \), \( \frac{1}{8} \), \( 1 \), \( \frac{1}{4} \). Wait, no, maybe I misread the table. Wait the user's table: let's re - check. The first row: \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, the original problem's table: the first row \( x=-2 \), \( f(x)=\frac{1}{8} \)? Wait no, the user's table (from the image): \( x=-2 \), \( f(x)=\frac{1}{8} \); \( x=-1 \), \( f(x)=\frac{1}{4} \); \( x = 0 \), \( f(x)=\frac{1}{2} \); \( x = 1 \), \( f(x)=1 \); \( x = 2 \), \( f(x)=2 \). Wait, but the initial value is \( f(0) \), because for \( f(x)=ab^x \), when \( x = 0 \), \( f(0)=a\times b^0=a \). So we look at \( x = 0 \) in the table. From the table, when \( x = 0 \), \( f(0)=\frac{1}{2} \)? Wait no, the options include \( \frac{1}{2} \) as the first option. Wait, let's confirm the table again. The user's table:

\( x=-2 \), \( f(x)=\frac{1}{8} \)

\( x=-1 \), \( f(x)=\frac{1}{4} \)

\( x = 0 \), \( f(x)=\frac{1}{2} \)

\( x = 1 \), \( f(x)=1 \)

\( x = 2 \), \( f(x)=2 \)

So when \( x = 0 \), \( f(0)=\frac{1}{2} \), which is the initial value (since \( f(x)=ab^x \), \( f(0)=a \)).

Answer:

\(\frac{1}{2}\) (the first option in the list of choices, which is labeled with \(\frac{1}{2}\))