Operations with Functions

Operations with Functions

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

Example 1: (f + g)(x)

If f(x) = 3x and g(x) = x^2 - 1, find the given function. (f + g)(x)

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

f(x) = 3x
g(x) = x2 - 1

So f(x) + g(x)
= [3x] + [x2 - 1].

Arrange the terms in descending order.

Then (given) = x2 + 3x - 1.

Example 2: (f - g)(x)

If f(x) = 3x and g(x) = x^2 - 1, find the given function. (f - g)(x)

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

f(x) = 3x
g(x) = x2 - 1

So f(x) - g(x)
= [3x] - [x2 - 1].

-(x2 - 1) = -x2 + 1

So (given) = -x2 + 3x + 1.

Example 3: (fg)(x)

If f(x) = 3x and g(x) = x^2 - 1, find the given function. (f*g)(x)

(fg)(x) means f(x)⋅g(x).

f(x) = 3x
g(x) = x2 - 1

So f(x)⋅g(x)
= [3x]⋅[x2 - 1].

3xx2 = 3x3
3x⋅(-1) = -3x

So (given) = 3x3 - 3x.

Multiplying a polynomial by a monominal

Example 4: (f/g)(x)

If f(x) = 3x and g(x) = x^2 - 1, find the domain of the given function. (f/g)(x)

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

f(x) = 3x
g(x) = x2 - 1

So f(x)/g(x)
= [3x]/[x2 - 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 3x/(x2 - 1),
set [x2 - 1 = 0].

Excluded value

x2 - 1
= x2 - 12
= (x + 1)(x - 1)

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

Factor the difference of two squares (a2 - b2)

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.