You must have learned in your younger years about the operations of addition, subtraction, multiplication, and division on numbers from the set of real numbers. Well, now its time to move on to more general quantities. Algebra of Continuous Functions deals with the use of continuous functions in equations involving the various binary operations you have studied so. We’ll also talk about a composition rule that might not be familiar to you but is very important for future applications. So let’s learn more about them!

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## Algebra of Continuous Functions

Suppose f(x) and g(x) are two continuous functions at the point x = a. Then we have the following rules:

## Addition and Subtraction Rules

- \({ \text{f(x) + g(x) is continuous at x = a}} \)
- \({ \text{f(x) – g(x) is continuous at x = a}} \)

**Proof:** We have to check for the continuity of (f(x) + g(x)) at x = a. Therefore, we’ll need to check for the three conditions of continuity to be satisfied. Note that since the functions f(x) and g(x) are continuous at x = a. the three conditions for continuity would automatically be satisfied for them i.e.

- f(a) and g(a) are defined
- Lim
_{x→a}f(x) = f(a) = k_{1}(say) and Lim_{x→a}g(x) = g(a) = k_{2}(say)

**Browse more Topics under Continuity And Differentiability**

- Continuity
- Algebra of Derivatives
- Derivatives of Composite Functions
- Derivatives of Functions in Parametric Forms
- Derivatives of Implicit Functions
- Derivatives of Inverse Trigonometric Functions
- Exponential and Logarithmic Functions
- Logarithmic Differentiation
- Mean Value Theorem
- Second Order Derivatives

Using them, we can get:

–> [f(a) + g(a)] is clearly defined at x = a because both f(a) and g(a) are defined.

–> Using the Summation Law of limits i.e. The limit of a sum is the sum of the limits; we can get:

Lim_{x→a}[f(x) + g(x)] = Lim_{x→a}f(x) + Lim_{x→a}g(x) = k_{1} + k_{2} (here)

–> f(a) + g(a) = k_{1} + k_{2} = Lim_{x→a}[f(x) + g(x)]

Hence, the function [f(x) + g(x)] is continuous at x = a. The proof for subtraction rule is similar to the proof for the addition rule (just replace the + sign with a – sign).

## Multiplication and Division Rules

- \({ f(x) \times{ g(x)} \text{ is continuous at x = a}} \)
- \({ \text{f(x)/g(x) is continuous at x = a; provided g(a) }\neq{ 0}} \)

**Proof: **Using the Product Law of limits i.e. The limit of a product is the product of the limits; we can get:

Lim_{x→a}[f(x) × g(x)] = Lim_{x→a}f(x) × Lim_{x→a}g(x) = k_{1} × k_{2} (here)

Using the Quotient Law of limits i.e. The limit of a quotient is the quotient of the limits; we can get:

Lim_{x→a}[f(x)/g(x)] = Lim_{x→a}f(x)/Lim_{x→a}g(x) = k_{1}/k_{2} (here, provided k_{2} ≠ 0)

Then the proofs will follow similarly. Now take a look at solved question (1) showing the applicability of one of these rules.

### Composition Rule

$$ { \text{f(g(x)) and g(f(x)) are continuous at x = a }} $$

We can prove it in a very straightforward fashion by showing that the three conditions of continuity hold true in this case. For an application look at solved question (2).

## Solved Examples for You

**Question 1**: Is the function f(x) = sin x + x^{3} + 5 continuous at x = 0?

**Solution:** The given function can be re-written as:

f(x) = (sin x) + (x^{3} + 5)

As you can see, we have broken down the function f(x) into a sum of two independent functions: g_{1}(x) = sin x; and g_{2}(x) = x^{3} + 5, out of which g_{1}(x) is clearly a trigonometric function and g2(x) is a polynomial function.

We know that trigonometric functions like sin x and polynomial functions are continuous at all points in their domain i.e. x ∈ the set of real numbers. Thus, f(x) is actually a function formed by the sum of two continuous functions. At point x = 0, sin x and (x^{3} + 5) are continuous. Using the Addition Rule then, we get

f(x) = (sin x) + (x^{3} + 5) is continuous at x = 0.

**Question 2:** Is the function f(x) = sin(x^{3} + 5) continuous at x = 0?

**Solution: **Without going into the trouble of showing the validity of the conditions of continuity here, one can see that this function is formed by the composition of two continuous functions: g_{1}(x) = sin x and g_{2}(x) = (x^{3} + 5). $${f(x) = g_1(g_2(x))} $$ Thus, by the composition rule, f(x) is continuous at x = 0.

This concludes our discussion on this topic of the algebra of continuous functions.

**Question 3: How can we get to know if a function is continuous?**

**Answer:** To confirm that a function is continuous, follow these steps:

- The function ‘f(c)’ should be defined. The function must be at an ‘x’ value (c), this means we can’t have a hole in this function.
- The limit of this function as ‘x’ approaches the value ‘C’ need to exist.
- The value of the function at ‘C’ and the limit as ‘x’ approaches ‘C’ should be similar.

**Question 4: Give an example of the continuous function.**

**Answer:** When a function is continuous in nature within its domain, then it is a continuous function. For instance, g(x) does not contain the value ‘x = 1’, so it is continuous in nature.

**Question 5: Are all continuous functions differentiable?**

**Answer:** Any differentiable function can be continuous at all points in its domain. For instance, a function having a bend, cusp, or a vertical tangent might be continuous in nature, but it fails to be differentiable at the place of the anomaly.

**Question 6: Are logarithmic functions continuous?**

**Answer:** The logarithmic functions are only defined for positive and real numbers. These are continuous at all points of the definition.