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

Given: f(x) = [x ], 0 < x < 3 Rf ' (1) = ^ lim _ h 0 f(1 + h)-f(1) h = ^ lim _ h 0 |1 + h |-1 h = ^ lim _ h 0 1-1 h = ^ lim _ h 0 0 h =0 And Lf ' (1) = ^ lim _ h 0 f(1 - h)-f(1) -h = ^ lim _ h 0 |1 - h| - 1 -h = ^ lim _ h 0 |1-h|-1 h = ^ lim _ h 0 0-1 -h = Since R f ' (1) L f ' (1) Therefore, f(x) = [x] is not differentiable at x =1. Similarly f(x) = [x] is not differentiable at x = 2.

Prove that the greatest integer function defined by f(x) = [x ], …

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Question

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

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Answer

Given: $\text{f(x)} = [\text{x} ], 0 < \text{x} < 3$
$\text{R}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 + h)}-\text{f}(1)}{\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 + \text{h} |-1}{\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{1-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0}{\text{h}}=0$
$\text{And}\ \text{L}\text{f}{'}(1) = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{\text{f(1 - h)}-\text{f}(1)}{-\text{h}} = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1 - \text{h}| - 1}{-\text{h}}$$ = ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{|1-\text{h}|-1}{\text{h}}= ^{\ \ \text{lim}}_{\text{h}\rightarrow\text{0}}\frac{0-1}{-\text{h}}=\infty$
Since $\text{R }\text{f}{'}(1)\neq \text{L}{\text{f}}{'}(1)$
Therefore, f(x) = [x] is not differentiable at x =1.
Similarly f(x) = [x] is not differentiable at x = 2.

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