Algebraic operations on continuous functions involve different rules that deal with their application in algebraic equations involving various binary operations. As the name states, a continuous function is a type of function with a continuous graph, i.e., one without any break or jumps. We can draw these types of graphs without lifting our pencils. A continuous function is a very important aspect of calculus as only a continuous function is differentiable. If there is a little break or jump in the graph, the function is said to be non-continuous. All polynomial functions on the set of real numbers are continuous. The absolute value for |x| and exponential functions are continuous at real numbers. Let’ us study the Algebraic operations on continuous functions and their rules in brief.
f + g: continuous at x = x0,
f – g: continuous at x = x0
To prove the continuity of (f(x) + g(x)) at x = a, we have to verify for three different conditions. As we know, f(x) and g(x) are continued functions (at x=a),
We know that f(a) and g(a) are defined
Assume,xaf(x) = f(a) = k1 and xag(x) = g (a) = k2
We get by using them:
= [f(a) + g(a)] is clearly defined at x = a {as both f(a) and g(a) are defined}.
= According to the Summation Law of limits that says that the limit of a sum is always equal to the sum of the limits, we get:
xa [f(x) + g(x)]
= xaf(x) + xag(x)
= k1 + k2 (here)
=> f(a) + g(a) = k1 + k2 = xa[f(x) + g(x)]
Therefore, we can conclude that the function [f(x) + g(x)] is continuous at x = a. (for the subtraction rule, we just need to replace signs of addition ‘+’ by subtraction ‘-‘)
f(x)×g(x): continuous at x = a
We can prove the multiplication rule by using the merchandise Law of limits.
The merchandise law of limit states that the limit of a product is the product of the limits.
On applying it we get:
xa [f(x) × g(x)]
= xa f(x) × xa g(x)
= k1 × k2 (here)
f(x)/g(x): continuous at x = a
and g(a) ≠0
We can prove the division rule by using the Quotient Law of limit.
The quotient law of limits states that the limit of a quotient is the quotient of the limit.
On applying it we will get:
xa [f(x)/g(x)]
= xa f(x)/xa g(x)
= k1/k2 (here, given k2 ≠ 0)
f(g(x)) and g(f(x)): continuous at x = a
Q 1- Determine if the function f(x) = sin(x3 + 5) is continuous (at x = 0)?
Solution:
By looking at the question, we can say that this function is the composition of two continuous functions:
g1(x) = sin x and
g2(x) = (x3 + 5)
f(x)=g1[g2(x)]
Thus, by the composition rule as stated above,
f(x) is continuous at x = 0.
Q 2- A function is given as f(x) = 3x – 7. Find whether it is a continuous function at x = 7.
Solution:
According to the question,
f(x) = 3x – 7, where x = 7 (a)
Finding thexaf(x) and f(a).
xaf(x)
=x7 (3x – 7)
= 3(7) – 7
= 21 – 7
= 14.
f(a) = f(7)
= 3(7) – 7
= 21 – 7
= 14.
So we can conclude that,xaf(x) = f(a).
Therefore, f(x) = 3x – 7 is continuous at x = 7.
Algebraic Operations on Continuous Functions are the functions that deal with the application of continuous functions in the algebraic equation that involves various binary operations. As the name says, a continuous function is a type of function with a continuous graph without any break or jumps. We can draw these types of graphs without lifting up our pencils. A continuous function is a very important aspect of calculus as only a continuous function is differentiable. All polynomial functions on the set of real numbers are continuous. The absolute value for |x| and exponential functions are continuous at real numbers. There are rules for addition, subtraction, multiplication, and division for a continued fraction as stated above.
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