Problems On Differentiability
Problem 1: Prove that the greatest integer function defined by f(x) = [x] , 0 < x < 3 is not differentiable at x = 1 and x = 2.
Solution:
As question given f(x) = [x] where x is greater than 0 and also less than 3. So we have to check the function is differentiable at point x =1 and at x = 2 or not. To check the differentiability of function, as we discussed above in Differentiation that LHD at(x=a) = RHD at (x=a) which means,
Lf’ at (x = a) = Rf’ at (x = a) if they are not equal after solving and putting the value of a in place of x then our function should not differentiable and if they both comes equal then we can say that the function is differentiable at x = a, we have to solve for two points x = 1 and x = 2. Now, let’s solve for x = 1
f(x) = [x]
Put x = 1 + h
Rf’ = limh -> 0 f(1 + h) – f(1)
= limh -> 0 [1 + h] – [1]
Since [h + 1] = 1
= limh -> 0 (1 – 1) / h = 0
Lf'(1) = limh -> 0 [f(1 – h) – f(1)] / -h
= limh -> 0 ( [1 – h] – [1] ) / -h
Since [1 – h] = 0
= limh -> 0 (0 – 1) / -h
= -1 / -0
= ∞
From the above solution it is seen that Rf’ ≠ Lf’, so function f(x) = [x] is not differentiable at x = 1. Now, let’s check for x = 2. As we solved for x = 1 in the same we are going to solve for x = 2. Condition should be the same we have to check that, Lf’ at (x = 2) = Rf’ at (x = 2) or not if they are equal then our function is differentiable at x = 2 and if they are not equal our function is not differentiable at x = 2. So, let’s solve.
f(x) = [x]
Differentiability at x=2
Put x = 2+h
Rf'(1) = limh -> 0 f(2 + h) – f(2)
= limh -> 0 ([2 + h] – [2]) / h
Since 2 + h = 2
= limh -> 0 (2 – 2) / h
= limh -> 0 0 / h
=0
Lf'(1) = limh -> 0 (f(2 – h) – f(2)) / -h
= limh -> 0 ([2 – h] – [2]) / -h
= limh -> 0 (1 – 2) / -h
Since [2 – h] =1
= -1 / -0
= ∞
From the above solution it is seen that Rf'(2) ≠ Lf'(2), so f(x) = [x] is not differentiable at x = [2]
Problem 2:
Show that the above function is not derivable at x = 0.
Solution:
As we know to check the differentiability we have to find out Lf’ and Rf’ then after comparing them we get to know that the function is differentiable at the given point or not. So let’s first find the Rf'(0).
Rf'(0) = limh -> 0 f(0 + h) – f(0)
= limh -> 0 (f(h) – f(0)) / h
= limh -> 0 h . [{(e(1 / h) – 1) / (e(1 / h) + 1) } – 0]/h
= limh -> 0 (e(1 / h) – 1) / (e(1 / h) + 1)
Multiply by e(-1 / h)
= limh -> 0 {1 – e(-1 / h) / 1 + e(-1 / h)}
= (1 – 0) / (1 + 0)
= 1
After solving we had find the value of Rf'(0) is 1. Now after this let’s find out the Lf'(0) and then we will check that the function is differentiable or not.
Lf'(0) = limh -> 0 { f(0 – h) – f(0) } / -h
= limh -> 0 -h . [{e(-1 / h) – 1 / e(-1 / h) + 1} – 0] / -h
= limh -> 0 { (e(-1 / h) – 1) / (e(-1 / h) + 1) }
= limh -> 0 { (1 – e(-∞)) / (1+e(-∞))}
= (0 – 1) / (0 + 1)
= -1
As we saw after solving Lf'(0) the value we get -1. Now checking if the function is differentiable or not, Rf'(0) ≠ Lf'(0) (-1≠1). Since Rf'(0) ≠ Lf'(0), so f(x) is not differentiable at x = 0.
Problem 3: A function is f(x) defined by
f(x) = 1 + x of x < 2
f(x) = 5 – x of x ≥ 2
If function f(x) differentiable at x = 2?
Solution:
So, for finding Lf'(2) we take the function f(x) = 1 = x, in the same way for finding Rf'(2) we take the function f(x) = (5 – x). Let’s find out Lf'(2) and Rf'(2)
Lf'(2) = limh -> 0 {f(2 – h) – f(2)} / -h
= limh -> 0 [[(2 – h) + 1] – [5 – 2]] / -h
= limh -> 0 (3 – h – 3) / -h
= limh -> 0 -h / -h
= 1
Rf'(2) = limh -> 0 {f(2 + h) – f(2)} / h
= limh -> 0 [[5 – (2 + h)] – 3] / h = limh-> 0 [5-2-h – 3] / h
=limh -> 0 -h / h
= -1
In the first line Lf'(2) after putting in the formula, for f(2) we are putting second function (5 – x). After solving the Lf'(2) we get the value 1. For calculating Rf'(2) we are using the second function 5-x and putting in the formula of Rf’, on solving the Rf'(2) we get the value -1.
Since, Rf'(2) ≠ Lf'(2) so we can say the function f(x) is not differentiable at x = 2.
Differentiability of a Function | Class 12 Maths
Continuity or continuous which means, “a function is continuous at its domain if its graph is a curve without breaks or jumps”. A function is continuous at a point in its domain if its graph does not have breaks or jumps in the immediate neighborhood of the point.
Continuity at a Point:
A function f(x) is said to be continuous at a point x = a of its domain, if
(LHL) = (RHL) = f(a) or lim f(x) = f(a)
where
- (LHL)x = a = limx -> a- f(x) and
- (RHL)x = a = limx -> a+ f(x)
Note: To evaluate LHL and RHL of a function f(x) at x = a, put x – h and x + h respectively, where h -> 0
Discontinuity at a Point:
If f(x) is not continuous at a point x = a, then it is discontinuous at x = a.
There are various types of discontinuity:
- Removable discontinuity: If limx -> a- f(x) = limx -> a+ f(x) ≠ f(a)
- Discontinuity of the first kind: If limx -> a+ f(x) ≠ limx -> a+ f(x)
- Discontinuity of the second kind: If limx -> a- f(x) or limx -> a+ f(x) both do not exist