A function f is defined on [-3,3] asf(x) = be | vedclass

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

Question

(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \ge

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) For $x \in [-2, 2]$,$f(x) = \min \{|x|, 2-x^2\}$.Solving $|x| = 2-x^2$,we get $x^2 + |x| - 2 = 0$,which gives $(|x|+2)(|x|-1) = 0$. Since $|x| \geq 0$,we have $|x| = 1$,so $x = \pm 1$.Thus,$f(x) = |x|$ for $x \in [-1, 1]$ and $f(x) = 2-x^2$ for $x \in [-2, -1) \cup (1, 2]$.Points of non-differentiability in $(-2, 2)$ are $x = -1, 0, 1$ (due to $|x|$ and the intersection of curves).For $2 For $x \in (2, 3]$,$f(x) = [x] = 2$ for $x \in (2, 3)$ and $f(3) = 3$. This is discontinuous at $x = 3$ (not in $(-3, 3)$) and $x = 2$ (jump discontinuity).For $x \in [-3, -2)$,$f(x) = [|x|] = [|x|]$. For $x \in (-3, -2)$,$f(x) = 2$. At $x = -2$,$f(-2) = 2$. At $x = -3$,$f(-3) = 3$.Checking points: $x = -2, -1, 0, 1, 2$. At these $5$ points,the function is either discontinuous or has a sharp corner.Therefore,the number of points of non-differentiability is $5$.

$A$ function $f$ is defined on $[-3,3]$ as
$f(x) = \begin{cases} \min \{|x|, 2-x^{2}\} & , -2 \leq x \leq 2 \\ [|x|] & , 2 < |x| \leq 3 \end{cases}$
where $[x]$ denotes the greatest integer $\leq x$. The number of points,where $f$ is not differentiable in $(-3,3)$ is

Similar Questions

Let $f(x) = \begin{cases} 1, & \forall x < 0 \\ 1 + \sin x, & \forall 0 \le x \le \pi/2 \end{cases}$,then what is the value of $f'(x)$ at $x = 0$?

Let $g(x) = x \cdot f(x)$,where $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$. Discuss the differentiability of $g$ at $x = 0$.

If $f(x) = \begin{cases} x + 2, & -1 < x < 3 \\ 5, & x = 3 \\ 8 - x, & x > 3 \end{cases}$,then at $x = 3$,$f'(x) = $

Prove that the greatest integer function defined by $f(x) = [x], 0 < x < 3$ is not differentiable at $x = 1$ and $x = 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ function $f$ is defined on $[-3,3]$ as$f(x) = \begin{cases} \min \{|x|, 2-x^{2}\} & , -2 \leq x \leq 2 \\ [|x|] & , 2 < |x| \leq 3 \end{cases}$where $[x]$ denotes the greatest integer $\leq x$. The number of points,where $f$ is not differentiable in $(-3,3)$ is