# Operations with Functions

How to solve the operations with functions: examples and their solutions.

## Example 1: (*f* + *g*)(*x*)

(*f* + *g*)(*x*) means *f*(*x*) + *g*(*x*).

*f*(*x*) = 3*x**g*(*x*) = *x*^{2} - 1

So *f*(*x*) + *g*(*x*)

= [3*x*] + [*x*^{2} - 1].

Arrange the terms in descending order.

Then (given) = *x*^{2} + 3*x* - 1.

## Example 2: (*f* - *g*)(*x*)

(*f* - *g*)(*x*) means *f*(*x*) - *g*(*x*).

*f*(*x*) = 3*x**g*(*x*) = *x*^{2} - 1

So *f*(*x*) - *g*(*x*)

= [3*x*] - [*x*^{2} - 1].

-(*x*^{2} - 1) = -*x*^{2} + 1

So (given) = -*x*^{2} + 3*x* + 1.

## Example 3: (*f*⋅*g*)(*x*)

(*f*⋅*g*)(*x*) means *f*(*x*)⋅*g*(*x*).

*f*(*x*) = 3*x**g*(*x*) = *x*^{2} - 1

So *f*(*x*)⋅*g*(*x*)

= [3*x*]⋅[*x*^{2} - 1].

3*x*⋅*x*^{2} = 3*x*^{3}

3*x*⋅(-1) = -3*x*

So (given) = 3*x*^{3} - 3*x*.

Multiplying a polynomial by a monominal

## Example 4: (*f*/*g*)(*x*)

(*f*/*g*)(*x*) means *f*(*x*)/*g*(*x*).

*f*(*x*) = 3*x**g*(*x*) = *x*^{2} - 1

So *f*(*x*)/*g*(*x*)

= [3*x*]/[*x*^{2} - 1].

Recall that

the denominator of a fraction cannot be 0.

And the *x* values

that make the denominator 0

are the excluded values.

So *x* cannot be the excluded values.

So, to find the excluded values of 3*x*/(*x*^{2} - 1),

set [*x*^{2} - 1 = 0].

Excluded value

*x*^{2} - 1

= *x*^{2} - 1^{2}

= (*x* + 1)(*x* - 1)

So (*x* + 1)(*x* - 1) = 0.

Factor the difference of two squares (*a*^{2} - *b*^{2})

Solve (*x* + 1)(*x* - 1) = 0.

1) *x* + 1 = 0

So *x* = -1.

2) *x* - 1 = 0

So *x* = 1.

So *x* = ±1.

These are the excluded values.

Solving a quadratic equation by factoring

[*x* = ±1] are the excluded values:

these should be excluded.

So [*x* ≠ ±1] is the answer.