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)

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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:

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