Answer:
Remember the property:
a^-1 = (1/a)^1
and:
(a/b)^n = (a^n)/(b^n)
A table for a function like:
[tex]\left[\begin{array}{ccc}x&f(x)\\&\\&\\&\\&\end{array}\right][/tex]
Is just completed as:
[tex]\left[\begin{array}{ccc}x&f(x)\\x_1&f(x_1)\\x_2&f(x_2)\\x_3&f(x_3)\\x_4&f(x_4)\end{array}\right][/tex]
So, here we have:
y = f(x) = (1/6)^x
To complete the table, we need to find:
f(-1)
and
f(2)
So let's find these two values:
f(-1) = (1/6)^-1 = (6/1)^1 = 6
and the other value is:
f(2) = (1/6)^2 = 1/36
Then the complete table is:
[tex]\left[\begin{array}{ccc}x&f(x)\\-2&36\\-1&6\\0&1\\1&1/6\\2&1/36\\1&1/216\end{array}\right][/tex]
