Find the range of the following function:
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.
$f(x) = 2 - 3x$,where $x \in R$ and $x > 0$.
(D) Given the function $f(x) = 2 - 3x$ for $x > 0$.
Since $x > 0$,we multiply both sides by $3$:
$3x > 0$
Now,multiply by $-1$ (which reverses the inequality sign):
$-3x < 0$
Add $2$ to both sides:
$2 - 3x < 2 - 0$
$f(x) < 2$
Thus,the range of the function $f$ is the set of all real numbers less than $2$.
Therefore,the range of $f$ is $(-\infty, 2)$.
Since $x > 0$,we multiply both sides by $3$:
$3x > 0$
Now,multiply by $-1$ (which reverses the inequality sign):
$-3x < 0$
Add $2$ to both sides:
$2 - 3x < 2 - 0$
$f(x) < 2$
Thus,the range of the function $f$ is the set of all real numbers less than $2$.
Therefore,the range of $f$ is $(-\infty, 2)$.
Explore More
Similar Questions
Let $f(x) = \frac{\tan^n x}{\sum_{r=0}^{2n} \tan^r x}$,$n \in N$,where $x \in [0, \frac{\pi}{2})$.
Match the following functions with their respective ranges:
Function Range $A. f(x) = |x|$ $I. [0, \infty)$ $B. f(x) = x^2$ $II. \mathbb{R}$ $C. f(x) = x^3$ $III. [0, \infty)$ $D. f(x) = \text{sgn}(x)$ $IV. \{-1, 0, 1\}$
| Function | Range |
|---|---|
| $A. f(x) = |x|$ | $I. [0, \infty)$ |
| $B. f(x) = x^2$ | $II. \mathbb{R}$ |
| $C. f(x) = x^3$ | $III. [0, \infty)$ |
| $D. f(x) = \text{sgn}(x)$ | $IV. \{-1, 0, 1\}$ |
The range of the function $f(x) = \frac{x}{1 + |x|}, x \in R,$ is
DifficultAIEEE 2012
View SolutionThe domain of the function $y = \frac{1}{\sqrt{|x| - x}}$ is
Easy
View SolutionThe range of the function $f(x) = (\sin x)^{\sin x}$ defined on $(0, \pi)$ is
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo