Question
The greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = _________.
✓
Answer
The greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = 1.Solution:
The given function f is f(x) = [x], 0 < x < 2
It is known that a function f is differentiable at a point x = c in its domain if both
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ and $\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}(1+\text{h})-\text{f(1)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^-}\frac{0-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
Consider the right hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^+}\frac{1-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}0=0$
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1
The given function f is f(x) = [x], 0 < x < 2
It is known that a function f is differentiable at a point x = c in its domain if both
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ and $\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}\text{(c+h)}-\text{f(c)}}{\text{h}}$ are finite and equal.
To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^-}\frac{\text{f}(1+\text{h})-\text{f(1)}}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^-}\frac{0-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{-1}{\text{h}}=\infty$
Consider the right hand limit of f at x = 1
$\lim\limits_{\text{h}\rightarrow0^+}\frac{\text{f}(1+\text{h})-\text{f}(1)}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}\frac{[1+\text{h}]-[1]}{\text{h}}$
$\lim\limits_{\text{h}\rightarrow0^+}\frac{1-1}{\text{h}}=\lim\limits_{\text{h}\rightarrow0^+}0=0$
Since the left and right hand limits of f at x = 1 are not equal, f is not differentiable at x = 1
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