The set of all points where the function f(x) | vedclass

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

Question

(A) The function is given by $f(x) = 2x|x|$.We can rewrite this as a piecewise function:$f(x) = \begin{cases} 2x^2, & x \geq 0 \ -2x^2, & x < 0 \end{

Vedclass · Accepted Answer

$(-\infty, \infty)$

Vedclass · Answer

(A) The function is given by $f(x) = 2x|x|$.We can rewrite this as a piecewise function:$f(x) = \begin{cases} 2x^2, & x \geq 0 \ -2x^2, & x Since $2x^2$ and $-2x^2$ are polynomials,$f(x)$ is differentiable for all $x 
eq 0$. We only need to check the differentiability at $x = 0$.Left Hand Derivative $(LHD)$ at $x = 0$:$f'(0^-) = \lim_{h ightarrow 0^-} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^-} \frac{-2h^2 - 0}{h} = \lim_{h ightarrow 0^-} (-2h) = 0$.Right Hand Derivative $(RHD)$ at $x = 0$:$f'(0^+) = \lim_{h ightarrow 0^+} \frac{f(0+h) - f(0)}{h} = \lim_{h ightarrow 0^+} \frac{2h^2 - 0}{h} = \lim_{h ightarrow 0^+} (2h) = 0$.Since $LHD = RHD = 0$,the function $f(x)$ is differentiable at $x = 0$.Therefore,$f(x)$ is differentiable for all $x \in (-\infty, \infty)$.

The set of all points where the function $f(x) = 2x|x|$ is differentiable is

Similar Questions

If $f(x) = \operatorname{Max} \{3 - x, 3 + x, 6\}$ is not differentiable at $x = a$ and $x = b$,then $|a| + |b| =$

If $f(x)=|x|+|\sin x|$ for $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,then its left hand derivative at $x=0$ is

If $f(x) = \begin{cases} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$,then the correct statement is:

If $f(x) = \begin{cases} \frac{x - 1}{2x^2 - 7x + 5} & \text{for } x \neq 1 \\ -\frac{1}{3} & \text{for } x = 1 \end{cases}$,then $f'(1) = $

Let $K$ be the set of all real values of $x$ where the function $f(x) = \sin |x| - |x| + 2(x - \pi) \cos |x|$ is not differentiable. Then the set $K$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module