F^-1= positive or negative x square root
Answer/Step-by-step explanation:
✔️As a set of order pairs, the mapping can be represented as (x, y), where x = input and y = output.
Thus:
{(1, 6), (2, 6), (3, 6), (4, 8)}
✔️As a table we have input as x, and y as output, such as:
x | y
1 | 6
2 | 6
3 | 6
4 | 8
Our equation looks like:

This means that we will need to distribute. This means that we need to multiply everything inside the parenthesis by the number outside of the parenthesis. (<em>Note: you only distribute when variables are involved.)</em>

Simplify:

We cannot simplify any further, so we know that our equation, until we know what x equals, is
.
a. (f+g)(x) = 8x+2
b. (f-g)(x) = 4x-6
Step-by-step explanation:
Given

<u>a. (f+g)(x)</u>

<u>b. (f-g)(x)</u>

Hence,
a. (f+g)(x) = 8x+2
b. (f-g)(x) = 4x-6
Keywords: Functions
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