Continuity and Differentiability

Algebra of Continuous Functions

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

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
  • Limx→af(x) = f(a) = k1 (say) and Limx→ag(x) = g(a) = k2 (say)

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:

Limx→a[f(x) + g(x)] = Limx→af(x) + Limx→ag(x) = k1 + k2 (here)

–> f(a) + g(a) = k1 + k2 = Limx→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:

Limx→a[f(x) × g(x)] = Limx→af(x) × Limx→ag(x) = k1 × k2 (here)

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

Limx→a[f(x)/g(x)] = Limx→af(x)/Limx→ag(x) = k1/k2 (here, provided k2 ≠ 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 + x3 + 5 continuous at x = 0?

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

f(x) = (sin x) + (x3 + 5)

As you can see, we have broken down the function f(x) into a sum of two independent functions: g1(x) = sin x; and g2(x) = x3 + 5, out of which g1(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 (x3 + 5) are continuous. Using the Addition Rule then, we get

f(x) = (sin x) + (x3 + 5) is continuous at x = 0.

Question 2: Is the function f(x) = sin(x3 + 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: g1(x) = sin x and g2(x) = (x3 + 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.

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