Answer:
Remember the property:
a^-1 = (1/a)^1
and:
(a/b)^n = (a^n)/(b^n)
A table for a function like:
![\left[\begin{array}{ccc}x&f(x)\\&\\&\\&\\&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5C%26%5C%5C%26%5C%5C%26%5C%5C%26%5Cend%7Barray%7D%5Cright%5D)
Is just completed as:
![\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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5Cx_1%26f%28x_1%29%5C%5Cx_2%26f%28x_2%29%5C%5Cx_3%26f%28x_3%29%5C%5Cx_4%26f%28x_4%29%5Cend%7Barray%7D%5Cright%5D)
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:
![\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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%26f%28x%29%5C%5C-2%2636%5C%5C-1%266%5C%5C0%261%5C%5C1%261%2F6%5C%5C2%261%2F36%5C%5C1%261%2F216%5Cend%7Barray%7D%5Cright%5D)
<em>f(3) = 13</em>
- Step-by-step explanation:
<em>f(x) = 2x + 7</em>
<em>replace </em><em>x</em><em> with </em><em>3</em>
<em>f(3) = 2×3 + 7</em>
<em>= 6 + 7</em>
<em>= 13</em>
Think of a vertical number line. The positive values count down as we move down this number line. The negative numbers count up, so to speak, as we move down the number line. So we have -1, -2, -3, ... as we move down. This indicates that -5 is <u>not</u> smaller than -7.
Diagram is below.
