The document Relation Between Continuity & Differentiability JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.

All you need of JEE at this link: JEE

**E. Relation between Continuity & Differentiability**

If a function f is derivable at x then f is continuous at x.

If f(x) is derivable for every point of its domain, then it is continuous in that domain .

**The converse of the above result is not true :**

"If f is continuous at x, then f may or maynot be derivable at x"

The functions f(x) =^{ } & g(0) = 0 are continuous at x = 0 but not derivable at x = 0.

**Remark :**

(a) Let f'_{+}(a) = p & f'_(a) = q where p & q are finite then :

(i) p = q ⇒ f is derivable at x = a ⇒ f is continuous at x = a.

(ii) p ≠ q ⇒ f is not derivable at x = a but f is continuous at x = a

Differentiable ⇒ Continuous ; Non-differentiable Discontinuous

But Discontinuous ⇒ Non-differentiable .

**(b)** If a function f is not differentiable but is continuous at x = a it geometrically implies a sharp corner at x = a.

**Ex.15 If f(x) = ****, then find the value of k so that f(x) becomes continuous at x = 0. Hence, find all the points where the functions is non-differentiable.**

**Sol. **From the graph of f(x) it is clear that for the function to be continuous only possible value of k is 1.

Points of non-differentiability are x = 0, ±1.

**Ex.16 If f(x) = **** where [.] denotes the greatest integer function.**

**Discuss the continuity and differentiability of f(x) in [0, 2).**

**Sol.** Since 1 x - 1 < 1 then [x^{2} - 2x] = [(x - 1)^{2} - 1] = [(x - 1)^{2}] - 1 = 0 - 1 = -1

Graph of f(x) :

It is clear from the graph that f(x) is discontinuous at x = 1 and not differentiable at x =1/2,and x= 1

Further details are as follows : ⇒

Hence, which shows f(x) is not differentiable at x = 1/2 (as RHD = 4 and LHD = –4) and x = 1 (as RHD = 0 and LHD = 8). Therefore, f(x) is differentiable, for x ∈ [0, 2) - {1/2, 1}

**Ex.17 Suppose f (x) = **

**Sol.** For continuity at x = 1 we leave f (1^{–}) = 1 and f (1^{+}) = a + b + c

a + b + c = 1 ....(1)

for continuity of f ' (x) at x = 1 f ' (1^{–}) = 3; f ' (1^{+}) = 2a + b

hence 2a + b = 3 ....(2)

f '' (1–) = 6; f '' (1+) = 2a for continuity of f '' (x) 2a = 6 ⇒ a = 3

from (2), b = – 3 ; c = 1. Hence a = 3, b = – 3 ; c = 1

**Ex.18 Check the differentiability of the function f(x) = max {sin ^{-1} |sin x|, cos^{-1} |sin x|}.**

**Sol. **sin^{-1} |sin x| is periodic with period ⇒ sin^{-1} |sin x| =

Also cos^{-1} |sin x| = - sin^{-1} |sin x|

⇒ f(x) is not differentiable at

⇒ f(x) is not differentiable at

**Ex.19 Find the interval of values of k for which the function f(x) = |x ^{2} + (k - 1) |x| - k| is non differentiable at five points.**

**Sol. **

f(x) = |x^{2} + (k – 1) |x| – k| = |(|x| – 1) (|x| + k)|

Also f(x) is an even function and f(x) is not differentiable at five points.

So |(x – 1) (x + k)| is non differentiable for two positive values of x.

⇒ Both the roots of (x – 1) (x + k) = 0 are positive.

⇒ k < 0 ⇒ k ∈ (–∝, 0).

**Definition : **A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a,b) [or (a, ∝) or (–∝, a) or (– ∝, ∝)] if it is differentiable at every number in the interval.

** Derivability Over An Interval : **f(x) is said to be derivable over an interval if it is derivable at each & every point of the interval. f(x) is said to be derivable over the closed interval [a, b] if :

(i) for the points a and b, f'(a^{+}) & f'(b ^{-}) exist &

(ii) for any point c such that a < c < b, f'(c^{+}) & f'(c ^{-}) exist & are equal .

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!