Let f : (-1, 1) to mathbb{R} be a function de | vedclass

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

Question

(C) The function is defined as $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$.To find the points of non-differentiability,we look for the intersection points

Vedclass · Accepted Answer

three elements

Vedclass · Answer

(C) The function is defined as $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$.To find the points of non-differentiability,we look for the intersection points of the curves $y = -|x|$ and $y = -\sqrt{1 - x^2}$.Setting $-|x| = -\sqrt{1 - x^2}$,we get $|x| = \sqrt{1 - x^2}$.Squaring both sides,$x^2 = 1 - x^2$,which implies $2x^2 = 1$,so $x^2 = \frac{1}{2}$,giving $x = \pm \frac{1}{\sqrt{2}}$.At $x = 0$,the function $f(x) = \min\{0, -1\} = -1$,which is a sharp corner for the function $-|x|$ and the intersection of the two curves.Checking the graph,the function $f(x)$ is non-differentiable at $x = -\frac{1}{\sqrt{2}}$,$x = 0$,and $x = \frac{1}{\sqrt{2}}$.Thus,there are $3$ points where the function is not differentiable. Therefore,$K$ has exactly three elements.

Let $f : (-1, 1) \to \mathbb{R}$ be a function defined by $f(x) = \min\{-|x|, -\sqrt{1 - x^2}\}$. If $K$ is the set of all points at which $f$ is not differentiable,then $K$ has exactly

Similar Questions

Let $f$ be a differentiable function from $R$ to $R$ such that $|f(x) - f(y)| \le 2|x - y|^{\frac{3}{2}}$ for all $x, y \in R$. If $f(0) = 1$,then $\int_{0}^{1} f^2(x) dx$ is equal to

Which of the following functions is differentiable at $x = 0$?

Let $f(x) = \begin{cases} x^2 \left| \cos \frac{\pi}{x} \right|, & x \neq 0 \\ 0, & x=0 \end{cases}$,$x \in \mathbb{R}$,then $f$ is

If $f(x) = \begin{cases} A + Bx^2, & x < 1 \\ 3Ax - B + 2, & x \geqslant 1 \end{cases}$,then find $A$ and $B$ so that $f(x)$ is differentiable at $x = 1$.

Let $f(x) = \begin{cases} x + 1, & \text{when } x < 2 \\ 2x - 1, & \text{when } x \ge 2 \end{cases}$,then $f'(2) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module