If f(x) = begin{cases} x[x], & 0 le x

Question

(A) We analyze the function $f(x)$ in the given intervals:For $0 \le x < 1$,$[x] = 0$,so $f(x) = x(0) = 0$.For $1 \le x < 2$,$[x] = 1$,so $f(x) = x(1)

Vedclass · Accepted Answer

neither $f'(1)$ exists nor $f'(2)$ exists

Vedclass · Answer

(A) We analyze the function $f(x)$ in the given intervals:For $0 \le x For $1 \le x For $2 \le x For $3 \le x At $x=4$,$f(4) = (4-1)[4] = 3(4) = 12$.Check differentiability at $x=1$:$LHL = \lim_{x 	o 1^-} f(x) = 0$.$RHL = \lim_{x 	o 1^+} f(x) = 1$.Since $LHL 
eq RHL$,$f(x)$ is discontinuous at $x=1$,therefore $f'(1)$ does not exist.Check differentiability at $x=2$:$f(2) = (2-1)[2] = 1(2) = 2$.$LHD = \lim_{h 	o 0^-} \frac{f(2+h) - f(2)}{h} = \lim_{h 	o 0^-} \frac{(2+h) - 2}{h} = \lim_{h 	o 0^-} \frac{h}{h} = 1$.$RHD = \lim_{h 	o 0^+} \frac{f(2+h) - f(2)}{h} = \lim_{h 	o 0^+} \frac{(2(2+h)-2) - 2}{h} = \lim_{h 	o 0^+} \frac{2h}{h} = 2$.Since $LHD 
eq RHD$,$f'(2)$ does not exist.Thus,neither $f'(1)$ nor $f'(2)$ exists.

If $f(x) = \begin{cases} x[x], & 0 \le x < 2 \\ (x-1)[x], & 2 \le x \le 4 \end{cases}$,where $[.]$ denotes the greatest integer function,then:

Similar Questions

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 $f^{\prime}(1)$ is equal to:

$A$ 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

The function that is not differentiable at $x=1$ is

If $f(x) = \begin{cases} e^x + ax, & x < 0 \\ b(x - 1)^2, & x \ge 0 \end{cases}$ is differentiable at $x = 0$,then $(a, b)$ is

If $f(x) = \begin{cases} 2x+3, & x \leq 1 \\ ax^{2}+bx, & x > 1 \end{cases}$ is differentiable $\forall x \in R$,then $f(2) = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module