Question

Differentiate f(x) = | x - 2 | at x = 2​

Answer 1

Answer:

0

Step-by-step explanation:

f(x) = x-2

(if x =2)

2f = 2-2

2f = 0

f = 0

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = |x - 2|$

We know, By definition of Modulus function, we have

$\rm \: \begin{gathered}\begin{gathered}\bf\: |x| = \begin{cases} &\sf{ - x \: \: when \: x < 0} \\ \\ &\sf{ \: \: x \: \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

So, using this definition,

$\rm \: \begin{gathered}\begin{gathered}\bf\: |x - 2| = \begin{cases} &\sf{ - (x - 2) \: \: when \: x - 2< 0} \\ \\ &\sf{ \: \: x - 2 \: \: when \: x - 2 \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

$\rm \: \begin{gathered}\begin{gathered}\bf\: |x - 2| = \begin{cases} &\sf{ - x + 2 \: \: when \: x < 2} \\ \\ &\sf{ \: \: x - 2 \: \: when \: x \geqslant 2} \end{cases}\end{gathered}\end{gathered}$

So,

$\rm \: \begin{gathered}\begin{gathered}\bf\: f(x) = \begin{cases} &\sf{ - x + 2 \: \: when \: x < 2} \\ \\ &\sf{ \: \: x - 2 \: \: when \: x \geqslant 2} \end{cases}\end{gathered}\end{gathered}$

Now, to find derivative of f(x) = | x - 2 | at x = 2, we have to check the differentiability of function at x = 2.

Consider, Left Hand derivative

$\rm \: \displaystyle\lim_{x \to 2^-}\rm \frac{f(x) - f(2)}{x - 2}$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\lim_{x \to 2^-}\rm \frac{ - x + 2 - 0}{x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2^-}\rm \frac{ - x + 2}{x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2^-}\rm \frac{ - (x - 2)}{x - 2}$

$\rm \: =  \: - \: 1$

$\rm\implies \:\boxed{\sf{  \:\displaystyle\lim_{x \to 2^ - }\rm \frac{f(x) - f(2)}{x - 2} \: = \: - \: 1 \: \: }}$

Now, Consider Right Hand Derivative,

$\rm \: \displaystyle\lim_{x \to 2^ + }\rm \frac{f(x) - f(2)}{x - 2}$

So, on substituting the values, we get

$\rm \: = \: \displaystyle\lim_{x \to 2^ + }\rm \frac{x - 2 - 0}{x - 2}$

$\rm \: = \: \displaystyle\lim_{x \to 2^ + }\rm \frac{x - 2}{x - 2}$

$\rm \: =  \:1$

$\rm\implies \:\boxed{\sf{  \:\displaystyle\lim_{x \to 2^ + }\rm \frac{f(x) - f(2)}{x - 2} \: = \: 1 \: \: }}$

So, from above steps, we concluded that

$\rm \: \displaystyle\lim_{x \to 2^ + }\rm \frac{f(x) - f(2)}{x - 2} \: \ne \: \displaystyle\lim_{x \to 2^ - }\rm \frac{f(x) - f(2)}{x - 2}$

$\rm\implies \:f(x) \: is \: not \: differentiable \: at \: x \: = \: 2$

$\rule{190pt}{2pt}$

Concept Used :- Differentiability

A function f(x) is said to be differentiable at x = a, iff

$\rm \:\boxed{\sf{  \: \: \displaystyle\lim_{x \to a^ + }\rm \frac{f(x) - f(a)}{x - a} \: = \: \displaystyle\lim_{x \to a^ - }\rm \frac{f(x) - f(a)}{x - a} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$