The function $f$ is defined by $f(x) = \begin{cases} 1 - x, & x < 0 \\ 1, & x = 0 \\ x + 1, & x > 0 \end{cases}$ Draw the graph of $f(x)$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
The function $f(x)$ is defined in three parts:
$1$. For $x < 0$,$f(x) = 1 - x$. This represents a straight line. For example,$f(-1) = 2$,$f(-2) = 3$,$f(-3) = 4$.
$2$. For $x = 0$,$f(0) = 1$. This is a single point $(0, 1)$.
$3$. For $x > 0$,$f(x) = x + 1$. This represents a straight line. For example,$f(1) = 2$,$f(2) = 3$,$f(3) = 4$.
Combining these,the graph consists of two rays starting from $(0, 1)$ extending into the second and first quadrants respectively.

Explore More

Similar Questions

Find the minimum integral value of $\alpha$ for which the graph of $f(x) = ||x - 2| - \alpha| - 5$ has exactly four $x-$intercepts.

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2}+2[x+2]-7=0$ has

Let $g: (-\infty, \infty) \to (-\frac{\pi}{2}, \frac{\pi}{2})$ be defined by $g(x) = 2 \tan^{-1}(e^x) - \frac{\pi}{2}$. Then $g(x)$ is...

Difficult
View Solution

The number of elements in the set $\{x \in R : (|x|-3)|x+4|=6\}$ is equal to

The number of positive integral values of $K$ for which the equation $K = |x + |2x - 1|| - |x - |2x - 1||$ has exactly three real solutions is

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