Find the range of the function f(x) defined b | vedclass

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

Question

(B) To find the range, we analyze the function in three intervals:
$1$. For $x < -1$, $f(x) = 2x - 3$. As $x 	o -1^-$, $f(x) 	o 2(-1) - 3 = -5$.

Vedclass · Accepted Answer

$(-\infty, -5) \cup [0, 1] \cup (5, \infty)$

Vedclass · Answer

(B) To find the range, we analyze the function in three intervals: $1$. For $x < -1$, $f(x) = 2x - 3$. As $x o -1^-$, $f(x) o 2(-1) - 3 = -5$. Since $x < -1$, $f(x) < -5$. So, the range for this part is $(-\infty, -5)$. $2$. For $-1 \leq x \leq 1$, $f(x) = 1 - x^2$. The minimum value is at $x = -1$ or $x = 1$, which is $1 - (1)^2 = 0$. The maximum value is at $x = 0$, which is $1 - 0 = 1$. So, the range for this part is $[0, 1]$. $3$. For $x > 1$, $f(x) = 3x^2 + 2$. As $x o 1^+$, $f(x) o 3(1)^2 + 2 = 5$. Since $x > 1$, $f(x) > 5$. So, the range for this part is $(5, \infty)$. Combining these, the total range is $(-\infty, -5) \cup [0, 1] \cup (5, \infty)$.

Find the range of the function $f(x)$ defined by:
$f(x) = \begin{cases} 2x-3, & x < -1 \\ 1-x^2, & -1 \leq x \leq 1 \\ 3x^2+2, & x > 1 \end{cases}$

Similar Questions

Let $f(x) = \sqrt{\frac{x+1}{x+3}}$ and $g(x) = \sqrt{\frac{2-x}{x+3}}$ be two real-valued functions. Then the domain of $f/g$ is

If the domain of the greatest integer function is the set of real numbers,then the range will be the set of

Let $f$ be a function such that $f(x) = \sum_{r=1}^n [r + \cos(\frac{x}{r})]$,where $[.]$ denotes the greatest integer function and $x \in [0, \pi]$. Then the range of $f(x)$ is:

The least integer in the range of $f(x) = \sqrt{(x + 4)(1 - x)} - \log_2 x$ is

The range of the function $f(x) = x^2 + \frac{1}{x^2+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Find the range of the function $f(x)$ defined by: $f(x) = \begin{cases} 2x-3, & x < -1 \\ 1-x^2, & -1 \leq x \leq 1 \\ 3x^2+2, & x > 1 \end{cases}$