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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let $g(x) = 1 - x + |x|$ and $h(x) = |x|$ for all real $x$.
Then the composite function $(h \circ g)(x) = h(g(x)) = h(1 - x + |x|) = |1 - x + |x|| = f(x)$.
Since $h(x) = |x|$ is a continuous function for all real $x$,and $g(x) = 1 - x + |x|$ is the sum of a polynomial function $(1 - x)$ and the modulus function $(|x|)$,both of which are continuous,$g(x)$ is also continuous.
Since $f(x)$ is the composition of two continuous functions $h$ and $g$,$f(x)$ is a continuous function for all real $x$.

Explore More

Similar Questions

Find all the points of discontinuity of the greatest integer function defined by $f(x) = [x]$,where $[x]$ denotes the greatest integer less than or equal to $x$.

$f(x) = \left[ \frac{x^2 + 1}{x^2[|x|] + 1} \right]$ is discontinuous at (where $[.]$ denotes the greatest integer function):

If the function $f(x) = \begin{cases} \frac{\sqrt{2 + \cos x} - 1}{(\pi - x)^2}, & x \neq \pi \\ k, & x = \pi \end{cases}$ is continuous at $x = \pi$,then $k$ equals:

If $f(x) = \begin{cases} \frac{1 - \sin x}{\pi - 2x}, & x \neq \frac{\pi}{2} \\ \lambda, & x = \frac{\pi}{2} \end{cases}$ is continuous at $x = \frac{\pi}{2}$,then the value of $\lambda$ is:

Let $f(x) = \frac{g(x)}{h(x)}$,where $g$ and $h$ are continuous functions on the open interval $(a, b)$. Which of the following statements is true for $a < x < b$?

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