Algebra of Continuous Functions

Algebra of continuous functions is defined for the four arithmetic operations:

• Addition of continuous functions
• Subtraction of continuous functions
• Multiplication of continuous functions
• Division of continuous functions

If two functions are continuous at a point, then the algebraic operations between two functions are also continuous. Let us understand the algebra of continuous functions with the respective theorem and proof. Also, we will solve examples to understand the concept better.

Also, read: Limits And Continuity

Addition of Continuous Functions

Theorem: Let us say, f and g are two real functions that are continuous at a point ‘a’, where ‘a’ is a real number. Then the addition of the two functions f and g is also continuous at ‘a’.

Proof: Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

Now as per the theorem,

lim_{x → a} (f+g)(x) ⇒ lim _{x → c} [f(x) + g(x)]

⇒ lim_{x → c} f(x) + lim_{x → c} g(x)

⇒ f(a) + g(a)

⇒ (f + g)(a)

Therefore,

lim_{x → a} (f+g)(x) = (f + g)(c)

Hence, f+g is continuous at x = a.

Subtraction of Continuous Functions

Theorem: Let us say, f and g are two real functions that are continuous at a point ‘a’, where ‘a’ is a real number. Then the subtraction of the two functions f and g is also continuous at ‘a’.

Proof: Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

Now as per the theorem,

lim_{x → a} (f – g)(x) ⇒ lim _{x → c} [f(x) – g(x)]

⇒ lim_{x → c} f(x) – lim_{x → c} g(x)

⇒ f(a) – g(a)

⇒ (f – g)(a)

Therefore,

lim_{x → a} (f – g)(x) = (f – g)(c)

Hence, f – g is continuous at x = a.

Multiplication of Continuous Functions

Theorem: If f and g are two real functions that are continuous at a point ‘a’, where ‘a’ is a real number. Then the product of the two functions f and g is also continuous at ‘a’.

Proof: Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

So, the limit of product of two functions, f and g at x is given by:

lim_{x → a} (f . g)(x) ⇒ lim _{x → c} [f(x) . g(x)]

⇒ lim_{x → c} f(x) . lim_{x → c} g(x)

⇒ f(a) . g(a)

⇒ (f . g)(a)

Therefore,

lim_{x → a} (f . g)(x) = (f . g)(c)

Hence, f . g is continuous at x = a.

Division of Continuous Function

Theorem: Suppose, f and g are two real functions that are continuous at a point ‘a’, where ‘a’ is a real number. Then the division of the two functions f and g will remain continuous at ‘a’.

Proof: Given,

lim_x→a f(x) = f(a)

lim_{x→ a} g(x) = g(a)

Now as per the theorem,

lim_{x → a} (f+g)(x) ⇒ lim _{x → c} [f(x) ÷ g(x)]

⇒ lim_{x → c} f(x) ÷ lim_{x → c} g(x)

⇒ f(a) ÷ g(a)

⇒ (f ÷ g)(a)

Therefore,

lim_{x → a} (f ÷ g)(x) = (f ÷ g)(c)

Hence, f ÷ g is continuous at x = a.

Algebra of Composite Functions

Suppose f and g are real valued functions such that (f o g) is defined at a. If g is continuous at a and if f is continuous at g (a), then (f o g) is continuous at a.

This theorem states that:

Solved Example

Q.1: Prove that f(x) = tan x is a continuous function.

Solution: Given,

f(x) = tan x

Since, we know that, by trigonometry formulas,

Tan x = sin x/cos x

Thus,

f(x) = sin x/cos x

Now, this function is defined for all real numbers and cos x ≠ 0, x ≠ (2n+1)π/2.

Sinx and cos x both are continuous functions, therefore, tan x is also continuous, since it is the quotient of sinx/cosx.

Q.2: Prove that the function defined by f(x) = | cos x | is a continuous function.

Solution: Given,

f(x) = |cos x|

f(x) is the real function for all real numbers ‘x’ and the domain of f(x) is the real number

Let g(x) = cos x

And h(x) = |x|

g(x) and h(x) are cosine functions and modulus functions are continuous for all real numbers.

Now,

(goh) (x) = g (h(x)) = g(|x|) = cos |x| is a composite function, hence is a continuous function. But it is not equal to f(x).

Again,

(hog) (x) = h(g(x)) = g(cos x) = |cos x| = f(x) [Given]

Hence,

f(x) = |cos x| = hog (x) is a composition function of two continuous functions. Therefore, it is also a continuous function.