Operations on functions

Functions work exactly as one would expect when it comes to the four basic algebraic operations (addition, subtraction, multiplication, and division). The domain of the resulting combined function is restricted to the elements shared by the domains of the original functions when these operations combine functions. This article discusses the functions of maths.

Basic operations on functions 

The formulas for combining functions are listed below.

Adding, subtracting, multiplying, and dividing functions with overlapping domains is possible. If f(x) and g(x) are two functions, then the sum, difference, product, and quotient are defined as follows for any x in the domain of both functions.

Adding two functions:

(f+g)(x) = f(x) + g(x)

Note: we put the f+g inside the brackets () to work simultaneously on both of x. 

Example: f(x) = 2x+3 and g(x) = x²

(f+g)(x) = (2x+3) + (x²) = x²+2x+3

Sometimes we may require to join all these terms:

Example: v(x) = 5x+1, w(x) = 3x-2

(v+w)(x) = (5x+1) + (3x-2) = 8x-1

The only other thing you need to worry about is the domain. Domains are the set of numbers that go along with the function. 

Subtracting two terms

You can also subtract two functions:

(f-g)(x) = f(x) − g(x)

Example: f(x) = 2x+3 and g(x) = x²

(f-g)(x) = (2x+3) − (x²)

Multiplication of two functions

You can also multiply two functions:

(f·g)(x) = f(x) · g(x)

Example: f(x) = 2x+3 and g(x) = x²

(f·g)(x) = (2x+3)(x2) = 2×3 + 3×2

Division of two functions

And you can divide two functions:

(f/g)(x) = f(x) / g(x)

Example: f(x) = 2x+3 and g(x) = x²

(f/g)(x) = (2x+3)/x²

There is yet another method where you can join two functions. The composition of two functions is the fifth operation. You can say that the functions f (x) and g(x): becomes (fog)(x). It’s the same as f (g(x)). “f of g of x.” The principle is simple and easy. The value of g at x is taken first, next is the value off along with that value. Let’s look at an example to help: 

Let f (x) = 3x and g(x) = x + 3. Then, what will be (fog)(x)? 

(fog)(x) = f (g(x)) = 3(x + 3) = 3x + 9. 

(gof )(x) = g(f (x)) = 3x + 3.

Rules for functional operations

But a function has special rules:

• Functions must work for every possible input value. 

• And it should have one relationship for each input value

This can be said in one definition: 

Definition: A function can relate to each element of a set with exactly one value of another set (possibly a similar set).

Two important considerations: 

1. “…each element…” Here, it means that every element in X is related to some element in Y. You can say that the function that covers X (relates with every element of it). (But some elements of Y may not be related at all.)  

2. “…exactly one…” Here, this means that every function is valued. It will not give back several results or output for the same input. So “f(2) = 7 or 9” is not the right answer. 

Let’s understand some operations on functions via some examples: 

1. If y is a function of x in the following equation -x²+ y² -2 = 0?. Find if y is a function of x? 

Answer: The answer to this question is first you need to solve the given equation for y.

y = + or – √ (x 2 +2)

y is not a function of x because for one value of the independent variable x (input) gives more than one output. After the solution, you obtain two values for the dependent variable y (output).

2. Determine the domain of function f given in the equation f(x) = √ ( x² + 1 ) + 2. 

Answer: The equation given inside the square root is positive for any real value of x. So the domain of function f is the set of all real numbers.

3. If any function f is given by f(x) = |x – 1| + 3. Determine f(-9). 

Answer: You  need to evaluate f(-9) by substituting x by -9 in the formula: 

f(-9) = |-9 – 1| + 3 = 10 + 3 = 13

4. If f and g are functions, give equation f(x) = 2x + 2 and g(x) = √ (x + 3). Determine (f o g)(6).

Answer: (f o g) (6) can be  calculated as follows: 

(f o g)(6) = f(g(6))

Let us first evaluate g(6) which is  g(6) = √ (6 + 3) = 3

We can now substitute g(6) by 3 in f(g(6))

f(g(6)) = f(3)

Then evaluate f(3),

f(3) = 2×3 + 2 = 8 and (f o g)(6) = 8 

5. Determine all real values of x such that f(x) = 0 given that f is a function defined by equation f(x) = (x² + 2 x – 3) / (x – 1)

Answer: f(x) = 0 gives the following equation: 

(x² + 2 x – 3) / (x – 1) = 0

Then the denominator is not equal to 0 for values of x not equal to 1.

The above equation tells us (x²+ 2 x – 3) = 0

Solve the above equation by factoring (x – 1)(x + 3) = 0

The above equation has two solutions: 

x = 1 and x = -3.

x = 1 is not an answer to f(x) = 0 because it is a value of x that makes the denominator of f(x) equal to zero. So the only value of x that can be f(x) = 0 is -3.

Conclusion

In mathematics, a function can be a relationship between one variable (the independent variable) and another variable (the dependent variable).We learnt basic rules like addition,subtraction,multiplication and division of functions.We also took some examples on domain and range as well as composite functions to strengthen our concepts.