Prove that every rational function is continuous at every point in its domain.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) rational function $f$ is defined as $f(x) = \frac{p(x)}{q(x)}$,where $p(x)$ and $q(x)$ are polynomial functions and $q(x) \neq 0$.
The domain of $f$ is the set of all real numbers $x$ such that $q(x) \neq 0$.
We know that every polynomial function is continuous everywhere on the set of real numbers $\mathbb{R}$.
According to the algebra of continuous functions,if $p(x)$ and $q(x)$ are continuous functions,then their quotient $\frac{p(x)}{q(x)}$ is also continuous at all points where the denominator $q(x) \neq 0$.
Since $p(x)$ and $q(x)$ are polynomials,they are continuous everywhere. Therefore,the rational function $f(x) = \frac{p(x)}{q(x)}$ is continuous at every point in its domain.

Explore More

Similar Questions

Let $f : [a, b] \rightarrow [1, \infty)$ be a continuous function and let $g : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $g(x) = \begin{cases} 0 & \text{if } x < a \\ \int_a^x f(t) dt & \text{if } a \leq x \leq b \\ \int_a^b f(t) dt & \text{if } x > b \end{cases}$. Then:

If $f(x) = \begin{cases} \frac{1 - \cos x}{x}, & x \ne 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$,then $k = $

If $f(x) = \begin{cases} x, & \text{if } x \text{ is irrational} \\ 0, & \text{if } x \text{ is rational} \end{cases}$,then $f$ is

Is the function defined by $f(x) = x^{2} - \sin x + 5$ continuous at $x = \pi$?

If $f(x) = \begin{cases} \frac{x^4 - 16}{x - 2}, & x \neq 2 \\ 16, & x = 2 \end{cases}$,then:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo