The range of the function f(x) = frac{x + 2}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the function $f(x) = \frac{x + 2}{|x + 2|}$.We know that the absolute value function $|x + 2|$ is defined as:$|x + 2| = \begin{cases} x + 2,

Vedclass · Accepted Answer

$\{-1, 1\}$

Vedclass · Answer

(B) Given the function $f(x) = \frac{x + 2}{|x + 2|}$.We know that the absolute value function $|x + 2|$ is defined as:$|x + 2| = \begin{cases} x + 2, & 	ext{if } x + 2 > 0 	ext{ (i.e., } x > -2) \ -(x + 2), & 	ext{if } x + 2 For $x > -2$,$f(x) = \frac{x + 2}{x + 2} = 1$.For $x Note that the function is undefined at $x = -2$ because the denominator becomes zero.Therefore,the range of the function $f(x)$ is $\{-1, 1\}$.

The range of the function $f(x) = \frac{x + 2}{|x + 2|}$ is

Similar Questions

The domain of definition of the function $y(x)$ given by the equation $2^x+2^y=2$ is

Let $f(x) = \cos(\pi(|x| + 2[x]))$,where $[.]$ represents the greatest integer function. Then:

The set of all real values of $x$ such that $f(x) = \sqrt{\frac{[x]-1}{[x]^2-[x]-6}}$ is a real-valued function is

If the domain of the function $f(x) = \sqrt{\log_{0.6} (\left| \frac{2x-5}{x^2-4} \right|)}$ is $(-\infty, a] \cup \{b\} \cup [c, d) \cup (e, \infty)$,then the value of $a+b+c+d+e$ is ————

Range of the function $f(x) = 3 + 2^x + 4^x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module