Question

A function f(x) is defined as f(x) [x],
where r is the greatest integer not greater
than x; -2 < x < 2. Discuss its continuity
and differentiability at x = 1.
1.​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

• A function f(x) is defined as f(x) = [x], where [.] is the greatest integer not greater than x; -2 < x < 2. Discuss its continuity and differentiability at x = 1.

$\underline{\bf \bold{ \: Answer \:}}$

Let us first define the function, f(x) = [x], where -2 < x < 2.

$\begin{gathered}\begin{gathered}\bf \:f(x) = \begin{cases} &\sf{ - 2 \: \: when \: - 2 < x < - 1} \\ &\sf{ - 1 \: \: when \: - 1 \leqslant x < 0}\\ &\sf{0 \: \: \: \: when \: \: 0\leqslant x < 1}\\ &\sf{1 \: \: \: when \: \: 1 \leqslant x < 2} \end{cases}\end{gathered}\end{gathered}$

Basic Concepts :-

Definition of continuity :-

• A function f(x) is continuous at x = 'a' iff

$\rm :\longmapsto\:f(1) = \bf \:\lim_{x\to 1^-} = \bf \:\lim_{x\to 1^ + }$

Definition of Differentiability :-

• A function f (x) is said to be differentiable iff

$\rm :\longmapsto\:\bf \:\lim_{x\to a^-} \: \dfrac{f(x) - f(a)}{x - a} = \:\lim_{x\to a^ + } \: \dfrac{f(x) - f(a)}{x - a}$

Let's solve the problem now!!

Now,

• Continuity at x = 1

$\rm :\longmapsto\:f(1) = 1 - - (1)$

Now,

Consider

• Left Hand Limit

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-} \: f(x)$

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-}0$

$\rm :\longmapsto\:0 - - (2)$

• From equation (1) and (2), we concluded

$\rm :\longmapsto\:f(1) \: \ne \: \bf \:\lim_{x\to 1^-}$

$\bf :\implies\: \boxed{ \bf \: f(x) \: is \: not \: continuous \: at \: x \: = \: 1}$

Now,

• To check differentiability at x = 1

Consider,

• Left Hand Derivative

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-} \: \dfrac{f(x) - f(1)}{x - 1}$

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-} \: \dfrac{0 - 1}{x - 1}$

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-} \: \dfrac{ - 1}{x - 1}$

$\rm :\longmapsto\:\bf \:\lim_{x\to 1^-} \: \dfrac{1}{1 - x}$

$\tt \: Put \: x = 1- h, \: as \: x \to \: 1 , \: so \: h \to \: 0$

$\rm :\longmapsto\:\bf \:\lim_{h\to 0} \: \dfrac{1}{1 - (1 - h)}$

$\rm :\longmapsto\:\bf \:\lim_{h\to 0} \: \dfrac{1}{h}$

$\rm :\longmapsto\: \infty \:( which \: is \: not \: finite \: number)$

$\rm :\implies\: \boxed{ \bf \: f(x) \: is \: not \: differentiable \: at \: x \: = \: 1}$

Additional Information :-

If f(x) and g(x) are continuous at x=a, and if c is a constant, then

• f(x) + g(x) is continuous at x = a.

• f(x) − g(x) is continuous at x = a.

• cf(x) is continuous at x = a.

• f(x)g(x) is continuous at x = a

• f(x)/g(x) is continuous at x = a, provided that g(a) ≠ 0.

• Every differentiable function is always continuous but converse needn't to be true.