Question

Question #f332a

Answer 1

See a solution process below:

Explanation:

For each of these values of `x` we need to substitute the value of `x` for `x` in the function and calculate the result:

• `color(red)(x) = color(red)(-2)`

`f(color(red)(-2)) = 3^color(red)(-2)`

`f(color(red)(-2)) = 1/3^color(red)(2)`

`f(color(red)(-2)) = 1/9`

• `color(red)(x) = color(red)(-1)`

`f(color(red)(-1)) = 3^color(red)(-1)`

`f(color(red)(-1)) = 1/3^color(red)(1)`

`f(color(red)(-1)) = 1/3`

• `color(red)(x) = color(red)(0)`

`f(color(red)(0)) = 3^color(red)(0)`

`f(color(red)(0)) = 1`

• `color(red)(x) = color(red)(1)`

`f(color(red)(1)) = 3^color(red)(1)`

`f(color(red)(1)) = 3`

Now that you know the pattern, see if you can do:

• `color(red)(x) = color(red)(2)`