Let f(x) = begin{cases} x+1, & -1 leq x leq 0 | vedclass

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

Question

(C) The function is defined as $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \ -x, & 0 < x \leq 1 \end{cases}$.Check continuity at $x=0$:Left-hand li

Vedclass · Accepted Answer

$f(x)$ is discontinuous in $[-1,1]$ but still has the maximum and minimum value

Vedclass · Answer

(C) The function is defined as $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \ -x, & 0 Check continuity at $x=0$:Left-hand limit: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (x+1) = 1$.Right-hand limit: $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (-x) = 0$.Since $\lim_{x 	o 0^-} f(x) 
eq \lim_{x 	o 0^+} f(x)$, the function is discontinuous at $x=0$.Now, analyze the range of $f(x)$ on $[-1, 1]$:For $x \in [-1, 0]$, $f(x) = x+1$. The range is $[0, 1]$.For $x \in (0, 1]$, $f(x) = -x$. The range is $[-1, 0)$.Combining these, the range of $f(x)$ is $[-1, 0) \cup [0, 1] = [-1, 1]$.The maximum value is $1$ (attained at $x=0$) and the minimum value is $-1$ (attained at $x=1$).Thus, $f(x)$ is discontinuous in $[-1, 1]$ but still has a maximum and minimum value.

Let $f(x) = \begin{cases} x+1, & -1 \leq x \leq 0 \\ -x, & 0 < x \leq 1 \end{cases}$. Which of the following statements is true?

Similar Questions

The value of $f(0)$,so that the function $f(x) = \frac{(27 - 2x)^{1/3} - 3}{9 - 3(243 + 5x)^{1/5}}, (x \ne 0)$ is continuous,is given by

If the function $f(x) = \begin{cases} 5x - 4, & 0 < x \le 1 \\ 4x^2 + 3bx, & 1 < x < 2 \end{cases}$ is continuous at every point of its domain,then the value of $b$ is

If $f(x) = \frac{e^{2x} - (1 + 4x)^{1/2}}{\ln(1 - x^2)}$ for $x \neq 0$,then $f$ has

Consider $f(x) = [x] + \sqrt{\{x\}}$,where $[.]$ denotes the greatest integer function and $\{.\}$ denotes the fractional part function. Identify the correct statement.

Vedclass Test Series

Exam Paper Generator

Online Exam Module