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

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  • Suppose f is a real function and c is a point in its domain the derivative of f at c is defined by
  • $\lim_{h \to 0}\large\frac{f(c+h)-f(c)}{h}$
  • $f'(x)=\lim_{h \to 0}\large\frac{f(x+h)-f(x)}{h}$
Step 1:
At $x=1$
Substitute $x=1$
RHD=$\lim_{h \to 0}\large\frac{f(1+h)-f(1)}{h}$
$\qquad=\lim_{h \to 0}\large\frac{1-1}{h}$$=0$
Step 2:
LHD=$\lim_{h \to 0}\large\frac{f(1-h)-f(1)}{-h}$
$\qquad=\lim_{h \to 0}\large\frac{0-1}{h}$=not defined.
Therefore f is not differentiable at $x=1.$
Step 3:
At $x=3$
RHD=$\lim_{h \to 0}\large\frac{f(3+h)-f(1)}{h}$
$\qquad=\lim_{h \to 0}\large\frac{3-3}{h}$$=0$
Step 4:
LHD=$\lim_{h \to 0}\large\frac{f(3-h)-f(3)}{-h}$
$\qquad=\lim_{h \to 0}\large\frac{2-3}{h}$
$\Rightarrow \lim_{h\to 0}\large\frac{-1}{h}$=not defined.
$\Rightarrow$ RHD $\neq $LHD
Therefore f is not differentiable at $x=3$
answered May 8, 2013 by sreemathi.v
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