Q. 205.0( 1 Vote )

# Examine the differentiability of f, where f is defined by

Answer :

Given,

…(1)

We need to check whether f(x) is continuous and differentiable at x = 2

A function f(x) is said to be continuous at x = c if,

Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).

Mathematically we can represent it as-

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-

Finally, we can state that for a function to be differentiable at x = c

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 2 if -

∴ LHL =

⇒ LHL = {using equation 1}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1; [-4.65] = -4 ; [9] = 9

∵ [2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHL = (2-0) ×1

∴ LHL = 2 …(2)

Similarly,

RHL =

⇒ RHL = {using equation 1}

∴ RHL = (1+0)(2+0) = 2 …(3)

And, f(2) = (2-1)(2) = 2 …(4) {using equation 1}

From equation 2,3 and 4 we observe that:

∴ f(x) is continuous at x = 2. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory-

f(x) is differentiable at x = 2 if -

∴ LHD =

⇒ LHD = {using equation 1}

Note: As [.] represents greatest integer function which gives greatest integer less than the number inside [.].

E.g. [1.29] = 1 ; [-4.65] = -4 ; [9] = 9

∵ [2-h] is just less than 2 say 1.9999 so [1.999] = 1

⇒ LHD =

⇒ LHD =

∴ LHD = 1 …(5)

Now,

RHD =

⇒ RHD = {using equation 1}

⇒ RHD =

∴ RHD =

⇒ RHD = 0+3 = 3 …(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=2) ≠ (RHD at x = 2)

∴ f(x) is not differentiable at x = 2

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