Differentiability Of A Function

DIFFERENTIABILITY OF A FUNCTION AT A GIVEN POINT

Let f (x) be a real valued function defined on an open interval (a,b) where c $\in \left(a,b\right).$Then f (x) is said to be differentiable or derivable at x = c,

$iff\underset{x\to c}{\lim }\frac{f(x)-f(c)}{(x-c)}$ exists finitely.

This limit is called the derivative or differential coefficient of the function

f(x) at x = c, and is denoted by f'(c) or D f (c) or $\frac{d}{dx}$ (f (x))_{x = c}

$\iff$Thus, f (x) is differentiable at x = c

$\underset{x\to c}{\lim }\frac{f(x)-f(c)}{(x-c)}existsfinitely$

$\Rightarrow \underset{x\to c^{-}}{\lim }\frac{f(x)-f(c)}{(x-c)}$ = $\underset{x\to c^{+}}{\lim }\frac{f(x)-f(c)}{(x-c)}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{f(c-h)-f(c)}{(-h)}$= $\underset{h\to 0}{\lim }\frac{f(c+h)-f(c)}{(h)}$

Here $\underset{x\to c^{-}}{\lim }\frac{f(x)-f(c)}{(x-c)}$= $\underset{h\to 0}{\lim }\frac{f(c-h)-f(c)}{(-h)}$ is called the left hand derivative of f (x) at x = c and is denoted by f'$(c^{-})$ or LF'(c).

While , $\underset{x\to c^{+}}{\lim }\frac{f(x)-f(c)}{(x-c)}$= $\underset{h\to 0}{\lim }\frac{f(c+h)-f(c)}{(h)}$ is called the right hand derivative

of f (x) at x=c and is denoted by f' $(c^{+})$or Rf' (c).

Thus f (x) is differentiable at x = c.

Lf'(c) = Rf' (c)

If Lf' (c) $\ne$ Rf'(c) we say that f (x) is not differentiable at x = c.

Example -1: The set of triplets (a, b, c) of real numbers with a $\ne$ 0, for which the function

f(x)=$\{\begin{array}{c}xforx\le 1\\ a{x}^2+bx+cotherwise\end{array}$, is differentiable, is

(A) { ( a, 1- 2a, a) / a ÎR; a ¹ 0 }

(B) { ( a, 1- 2a, c) /a, c ÎR; a ¹ 0 }

(C) { (a, b, c)/ a, b, c ÎR; a + b+ c = 1 }

(D) { ( a, 1- 2a, 0) / a ÎR; a ¹ 0 }

Solution: (A) Given f is differentiable for all real x

$\Rightarrow$ f is continuous for all real x.

so, $\underset{x\to 1}{\lim }$f(x) = f(1) $\Rightarrow$ a + b + c =1 . . . . (1)

Also f'(x) =$\{\begin{array}{c}1forx\le 1\\ 2ax+botherwise\left(\sin cefiscontinuous\right)\end{array}$

f' (1+) = f'(1–) $\Rightarrow$ 1 = 2a + b $\Rightarrow$ b = – 2a + 1 . . . . (2)

since a, b, c $\in$ R and a $\ne$ 0, using (1) and (2)

$\Rightarrow$ c = a

DIFFERENTIABILITY IN AN INTERVAL

IN OPEN INTERVAL

A function f(x) defined on an open interval (a, b) is said to be differentiable or derivable in open interval (a, b) if it is differentiable at each point of (a, b)

IN CLOSE INTERVAL

A function f(x) defined on [a, b] is said to be differentiable or derivable at the end points a and b if it is differentiable from the right at a and from the left at b. In other words $\underset{x\to a^{+}}{\lim }\frac{f(x)-f(a)}{x-a}$ and $\underset{x\to b^{+}}{\lim }\frac{f(x)-f(b)}{x-b}$ both exist.

"If f is derivable in the open interval (a, b) and also at the end points a and b, then f is said to be derivable in the closed interval [a, b]".

For checking differentiability on a closed interval [a, b] we say,

"A function f is said to be differentiable function if it is differentiable at every point of its domain."

Example -2: Let f(x) = x³ – x² + x + 1

g(x) = \left\{ \begin{align} \max \,f\left( t \right),0\le t\le x,\,\,\,forx\le 1 \\ 3-x,1<x\le 2 \\ \end{align} \right.

Discuss the continuity and differentiability of g(x) in (0, 2).

Solution: f(x) = x³ – x² + x + 1

$\Rightarrow$ f'(x) = 3x² – 2x + 1 > 0 $\forall$ x

$\Rightarrow$ f(x) is an increasing function on [0, x]

Hence, g(x) = \left\{ \begin{align} {{x}^{3}}-{{x}^{2}}+x+1,0\le x\le 1 \\ 3-x,1<x\le 2 \\\end{align} \right.

Clearly g(x) is continuous at x = 1 and not differentiable at x = 1.