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Karnataka Board PUCPUC Science 2nd PUC Class 12

Algebra of Continuous Functions

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Estimated time: 7 minutes
CBSE: Class 12

Introduction

Just as the concept of limits follows specific algebraic rules, the continuity of functions does as well. Because continuity at a point is entirely determined by the limit of the function at that point, the algebra of continuous functions mirrors the algebra of limits.

These rules allow us to easily determine the continuity of complex functions by breaking them down into simpler, known continuous parts.

CBSE: Class 12

Algebra of Continuous Functions

If f and g are two real functions that are continuous at a real number x = c, then the following algebraic combinations are also continuous at x = c:

  • Addition: f + g

  • Subtraction: f - g

  • Multiplication: \[f \cdot g\]

  • Division: \[\frac{f}{g}\] (provided that \[g(c) \neq 0\])

  • Reciprocal of a Continuous Function: If g is continuous and g(x) ≠ 0, then \[ \frac{1}{g(x)} \]

  • Rational Functions:  \[ f(x) = \frac{p(x)}{q(x)} \]
    where \[p(x)\] and \[q(x)\] are polynomial functions and \[ q(x) \neq 0. \]is continuous wherever it is defined.

  • Trigonometric Functions: \[\sin x\] and \[\cos x\] are continuous for all real \[x\], while \[\tan x = \dfrac{\sin x}{\cos x}\] is continuous wherever \[\cos x \neq 0\], i.e. \[x \neq \dfrac{(2n+1)\pi}{2}\], \[n \in \mathbb{Z}\].
CBSE: Class 12

Theorem: Composition of Functions

If g is continuous at c, and f is continuous at g(c), then their composite function \[(f \circ g)\], defined as \[(f \circ g)(x) = f(g(x))\], is also continuous at c.

CBSE: Class 12

Example 1

Show that the function defined by f(x) = sin (x2) is a continuous function.

Solution: 

Let \[ g(x) = \sin x \]

and \[ h(x) = x^2. \]

Since both \[g\] and \[h\] are continuous functions,

\[ f(x) = (g \circ h)(x) = g(h(x)) = \sin(x^2) \]

is continuous for all real \[x\].

CBSE: Class 12

Example 2

Show that the function f defined by f(x) = |1 – x + |x||, where x is any real number, is a continuous function.

Solution:

Let \[ g(x) = 1 - x + |x| \] and \[ h(x) = |x|. \]

Since polynomial functions and \[|x|\] are continuous, 

\[ g(x) \] is continuous.

Also \[h\] is continuous.

Therefore, \[ f(x) = h(g(x)) \] is continuous.

CBSE: Class 12
Maharashtra State Board: Class 12

Key Points: Algebra of Continuous Functions

  •  If \[f\] and \[g\] are continuous at \[c\], then \[f+g\], \[f-g\], and \[fg\] are continuous at \[c\].
  • \[\dfrac{f}{g}\] is continuous provided 
    \[ g(c) \neq 0. \]
  • If \[f\] is continuous, then \[\lambda f\] is continuous.
  • If \[g\] is continuous and non-zero, then \[ \frac{1}{g} \] is continuous.
  • Rational functions are continuous on their domains.
  • sin x and cos x are continuous for all real \[x\].
  • \[\tan x\] is continuous wherever \[\cos x \neq 0\].
  •  If \[g\] is continuous at \[c\] and \[f\] is continuous at \[g(c)\], then \[ f \circ g \] is continuous at \[c\].

Test Yourself

Video Tutorials

We have provided more than 1 series of video tutorials for some topics to help you get a better understanding of the topic.

Series 1


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Series 3


Shaalaa.com | Continuity and Differentiability part 12 (Continuity composite function)

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Continuity and Differentiability part 12 (Continuity composite function) [00:06:40]
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