What are the types of discontinuities?

There are several types of discontinuities: removable (a hole in the graph), jump (a sudden change in value), and infinite (vertical asymptote).

What is the significance of continuity in calculus?

Continuity is crucial in calculus because many theorems, such as the Intermediate Value Theorem and the Extreme Value Theorem, require functions to be continuous.

What is the differentiability in a function?

Differentiability means a function has a well-defined tangent (slope) at each point. A function f(x) is differentiable at x = a if the derivative f'(a) exists.

How is differentiability related to continuity?

Differentiability implies continuity. If a function is differentiable at x = a, it is also continuous at x = a. However, a continuous function is not necessarily differentiable.

What is the geometric interpretation of differentiability?

Geometrically, differentiability means the graph of the function has a well-defined tangent line at a point, implying a smooth curve without sharp edges or vertical slopes at that point.

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Continuity and Discontinuity

Continuity in mathematics means a function has no interruptions at a point, meeting three conditions: defined at the point, limit exists, and limit equals the function's value. Discontinuity occurs when these...

Continuity

Continuity of a function is a fundamental concept in mathematics that describes the connectedness of the graph of the function. Formally, a...

Determinant of a Matrix

The determinant of a matrix is a single numerical value that provides important insights into the properties and behavior of the...

Determinants and Matrices

Determinants are special numbers calculated from square matrices. They help find areas,...

Differential Equations

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a Differential Equation...

Domain and Range of a Relation

The domain is all the possible input values, while the range is all the possible output values. Knowing how...

Inverse Trigonometric Functions

Inverse Trigonometric Functions are the inverse counterparts of the fundamental trigonometric functions: sine, cosine,...

Continuity and Differentiability

Continuity and differentiability are key concepts in calculus. Continuity means a function has no breaks in its graph. Differentiability means the function's slope can be defined at every point. A function must be continuous to be differentiable, but not vice versa. Understanding these concepts helps in analyzing functions and solving mathematical problems effectively.

1.0Continuity

Continuity of a function is a fundamental concept in mathematics that describes the smoothness of the graph of the function. Formally, a function f(x) is considered to be continuous at a point x = c if three conditions are met:

f(c) is defined, meaning that the function has a value at x = c.
The function has a defined limit as x approaches the value c. This means that as x gets arbitrarily close to c when the values of f(x) approach a specific finite value.
The value of the function at x = c coincides with the limit of the function at x = c, i.e.,

$limf_{x \to c} (x) = f (c)$ .

A function f is continuous at x = c if it is defined at x = c and the value of f(c) equals the limit of f(x) as x approaches c. If f is not continuous at c, then f is said to be discontinuous at c, and c is referred to as a point of discontinuity of f.

Also Read Continuity in Interval

2.0Differentiability

In Calculus, a function f(x) is considered differentiable at a point x = c if its derivative exists at that point. This entails the limit of the difference quotient $\frac{f ( x ) - f ( c )}{x - c}$ approaching a finite value as x approaches c. Graphically, differentiability implies a smooth, continuous curve with a well-defined tangent line at c, f(c).

The derivative of a function f at x = c is defined by the limit: $f^{'} (c) = lim_{h \to 0} \frac{f ( c + h ) - f ( c )}{h}$ provided this limit exists. The derivative of f at x = c is denoted by f'(c). The function defined by $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ wherever the limit exists, is called the derivative of f. The derivative of f is denoted by f'(x), $\frac{d}{dx} f (x)$ , or if y = f(x), by $\frac{dy}{dx} or y^{'}$ .The process of finding the derivative of a function is known as differentiation. We also say, "differentiate f(x) with respect to x " to mean finding f'(x).

3.0Algebra of Continuous Function

Suppose u and v be two real functions continuous at a real number c. Then

u + v is continuous at x = c.
u – v is continuous at x = c.
u.v is continuous at x = c.
$\frac{u}{v}$ is continuous at x = c, (provided v(c) ≠ 0).

Also Read Continuity and Discontinuity

4.0Algebra of Derivative Function

The following rules were established as a part of algebra of derivatives:

(u ± v) ' = u' ± v'
(u.v)' = u'v + uv' (Leibnitz or product rule)
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ , (Quotient rule)

5.0Continuity and Differentiability Solved Examples

Example 1: Discuss the continuity of $f (x) = ⎩ ⎨ ⎧ ∣ x + 1∣, 2 x + 3, x^{2} + 3, x^{3} - 15, x < - 2 - 2 \leq x < 0 0 \leq x < 3 x \geq 3$

Solution:

We write f(x) as $f (x) = ⎩ ⎨ ⎧ - x - 1, 2 x + 3, x^{2} + 3, x^{3} - 15, x < - 2 - 2 \leq x < 0 0 \leq x < 3 x \geq 3$

As we can see, f(x) is defined as a polynomial function in every interval (–∞, –2), [0, 3), [–2, 0), and [3, ∞). Consequently, it is continuous in each of these 4 open intervals. Thus, we check the continuity at

x = –2, 0, 3

At the point

$lim_{x \to - 2^{-}} f (x) = lim_{x \to - 2^{-}} (- x - 1) = + 2 - 1 = 1$

$lim_{x \to - 2^{+}} f (x) = lim_{x \to - 2^{+}} (2 x + 3) = 2 \cdot (- 2) + 3 = - 1$

Therefore, $lim_{x \to - 2} f (x)$ does not exist and hence f(x) is discontinuous at .

At the point

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (2 x + 3) = 3$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (x^{2} + 3) = 3 f (0) = 0^{2} + 3 = 3$

Therefore f(x) is continuous at x = 0 .

At the point x = 3

$lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} (x^{2} + 3) = 3^{2} + 3 = 12$

$lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} (x^{3} - 15) = 3^{3} - 15 = 12$

f(3)=3^3-15=12

Therefore, f(x) is continuous at x = 3 .

We find that f(x) is continuous at all points in $R$ except at x = –2

Example 2: If $f (x) = {sin \frac{p x}{2}, [x] x < 1 x \geq 1$ then find whether f(x) is continuous or not at x = 1 , where [.] denotes greatest integer function.

Solution:

$f (x) = {sin \frac{p x}{2}, x < 1 [x], x \geq 1$

For continuity at x = 1, we determine, $f (1) lim_{x \to 1^{-}} f (x) and lim_{x \to 1^{+}} f (x)$ .

Now, f(1)=[1]=1 $f (x) = lim_{x \to 1^{-}} sin \frac{p x}{2} = sin \frac{p}{2} = 1 and lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} [x] = 1$

so $f (1) = lim_{x \to 1^{-}} f (x) = f (x)$

∴ f(x) is continuous at

Example 3:

If $f (x) = \frac{x + 1}{x - 1} and g (x) = \frac{1}{x - 2}$ , then discuss the continuity of f(x), g(x) and fog(x) in its domain.

Solution:

$f (x) = \frac{x + 1}{x - 1}$

f(x) is a rational function it must be continuous in its domain and f is not defined at .

$g (x) = \frac{1}{x - 2}$

g(x)is also a rational function. It must be continuous in its domain and g is not defined at x = 2 .

Consider g(x) = 1

$\frac{1}{x - 2} = 1 \Rightarrow x = 3$

∴ fog (x)is continuous in its domain: R – {2, 3}

Example 4: If f(x)= $[x - x if if x \in R x \in / R$ find the points where f(x) is continuous.

Solution:

Let be the point at which f(x) is continuous.

⇒ $lim_{x \to a throuthetraional} f (x) = lim_{x \to a throughitrational} f (x)$

⇒ a = –a

⇒ a = 0 ⇒ function is continuous at x = 0.

Example 5: Determine the values of x for which the following functions fails to be continuous or differentiable

$f (x) = ⎩ ⎨ ⎧ (1 - x), (1 - x) (2 - x), (3 - x), x < 1 1 \leq x \leq 2 x > 2 .$ Justify your answer.

Solution:

By the given definition the function f is continuous and differentiable at all points except possibility at x = 1 and x = 2.

Check the differentiability at x = 1

$\Rightarrow q = LHD = lim_{h \to 0} \frac{f ( 1 - h ) - f ( 1 )}{- h} = lim_{h \to 0} \frac{1 - ( 1 - h ) - 0}{- h} = - 1$

$\Rightarrow p = RHD = lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h} = lim_{h \to 0} \frac{{ 1 - ( 1 + h )} { 2 - ( 1 + h ) - 0 }}{h} = - 1$

⇒ $q = p or L . H . S = R . H . S$

∴ Differentiable at x =1 ⇒ Continuous at x = 1 .

Check the differentiability at x = 2

⇒ $q = LHD = lim_{h \to 0} \frac{f ( 2 - h ) - f ( 2 )}{- h} = lim_{h \to 0} \frac{( 1 - 2 + h ) ( 2 - 2 + h ) - 0}{- h} = 1 = finite$

⇒ $p = R HD = lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{( 3 - 2 - h ) - 0}{h} \to \infty (n o t f ini t e) ∵ q \neq = p$ ∴ not differentiable at x = 2 .

Now we have to check the continuity at

$\Rightarrow LHL = lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (1 - x) (2 - x) = lim_{h \to 0} (1 - (2 - h)) (2 - (2 - h)) = 0$

$\Rightarrow RHL = lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (3 - x) = lim_{h \to 0} (3 - (2 + h)) = 1$

∴ LHL ≠ RHL

⇒ not continuous at x = 2 .

Example 6: f(x)=|| x|-1| is not differentiable at

(A) 0 (B) ±1,0 (C) 1 (D) ±1

Ans. (B)

Solution:

= ${∣ x ∣ - 1, - ∣ x ∣ + 1, ∣ x ∣ - 1 \geq 0 ∣ x ∣ - 1 < 0$

= ${∣ x ∣ - 1, - ∣ x ∣ + 1, x \leq - 1 or x \geq 1 - 1 < x < 1$

$= ⎩ ⎨ ⎧ - x - 1, x + 1, - x + 1, x - 1, x \leq - 1 - 1 < x < 0 0 \leq x < 1 x \geq 1$

From the graph. It is clear that f(x) is not differentiable at x = –1,0 and 1.

Example 7: For what triplets of real numbers (a,b,c) with a $\neq = 0$ the function

$f (x) = {x, a x^{2} + b x + c, x \leq 1 otherwise$

is differentiable for all real x ?

(A) ${(a, 1 - 2 a, a) ∣ a \in R, a \neq = 0} (B) {(a, 1 - 2 a, c) ∣ a, c \in R, a \neq = 0}$

(C) ${(a, b, c) ∣ a, b, c \in R, a + b + c = 1} (D) {(a, 1 - 2 a, 0) ∣ a \in R, a \neq = 0}$

Ans. (A)

Solution:

$f (x) = {x a x^{2} + b x + c,, x \leq 1 otherwise$

⇒ f(x) should be continuous at x=1

it gives a + b + c = 1

⇒ f(x)should be differentiable at x =1

it gives $2 a + b = 1 \Rightarrow b = 1 - 2 a$

⇒ $c = 1 - a - b = a$

Example 8: If $f (x) = ⎩ ⎨ ⎧ e^{- ∣ x ∣}, - e^{- ∣ x - 1∣} + e^{- 1} + 1, e^{- ∣ x - 2∣}, - 5 < x < 0 0 \leq x < 2 2 \leq x < 4$

Discuss the continuity and differentiability of f(x) in the interval (–5, 4).

Solution:

$f (x) = ⎩ ⎨ ⎧ e^{+ x} - e^{x - 1} + e^{- 1} + 1 - e^{- x + 1} + e^{- 1} + 1 e^{- x + 2} - 5 < x < 0 0 \leq x \leq 1 1 < x < 2 2 \leq x < 4$

Check the differentiability at x = 0

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 0 - h ) - f ( 0 )}{- h} = lim_{h \to 0} \frac{e ^{- h} - 1}{- h} = 1$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h} = lim_{h \to 0} \frac{- e ^{h - 1} + e ^{- 1} + 1 - 1}{h} = - e^{- 1}$

$\Rightarrow LHD \neq = RHD$

∴ Not differentiable at , but continuous at since LHD and RHD both are finite.

Check the differentiability at x = 1

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 1 - h ) - f ( 1 )}{- h} = lim_{h \to 0} \frac{- e ^{1 - h - 1} + e ^{- 1} + 1 - e ^{- 1}}{- h} = - 1$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h} = lim_{h \to 0} \frac{- e ^{1 - h - 1} + e ^{- 1} + 1 - e ^{- 1}}{h} = 1$

LHD ≠ RHD

∴ Not differentiable at x = 1 , but continuous at x = 1 since LHD and RHD both are finite.

Check the differentiability at x = 2

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 2 - h ) - f ( 2 )}{- h} = lim_{h \to 0} \frac{- e ^{- 2 + h + 1} + e ^{- 1} + 1 - 1}{- h} = lim_{h \to 0} \frac{- e ^{- 1} ( e ^{h} - 1 )}{- h} = e^{- 1}$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{e ^{- h} - 1}{h} = - 1$

LHD ≠ RHD

∴ Not differentiable at x = 2, but continuous at x = 2 since LHD & RHD both are finite.

Example 9: The function $f (x) = (x^{2} - 1) x^{2} - 3 x + 2 + cos (∣ x ∣)$ is NOT differentiable at:

(A) –1 (B) 0 (C) 1 (D) 2

Ans. (D)

Solution:

$f (x) = (x^{2} - 1) ∣ (x - 1) (x - 2) ∣$

Clearly not differentiable at x = 2

Also Read Differential Equation previous year questions with solutions

6.0Continuity and Differentiability Practice Problems

1. The function $f (x) = \frac{4 - x ^{2}}{4 x - x ^{3}}$ , is -

(A) discontinuous at only one point in its domain.

(B) discontinuous at two points in its domain.

(C) discontinuous at three points in its domain.

(D) continuous everywhere in its domain.

Ans: C

2. A function y = f(x) is defined as $f (x) = {k sin \frac{( x + 3 ) p}{6} \frac{3 - 11 - x}{x - 2} for for x \leq 2 x > 2$ . If f(x) is continuous at x = 2, then find the value of k .

Ans: $k = \frac{1}{3}$

3. A function is defined as follows

$f (x) = {x^{3}; x; x^{2} < 1 x^{2} \geq 1$

is

(A) Continuous but not differentiable at x = 1

(B) Discontinuous but differentiable at x = 1

(C) Discontinuous but not differentiable at x = 1

(D) Continuous but differentiable at x = 1

Ans: A

4. A function is defined as follows $f (x) = (x - 1) ∣ x - 1∣$ .

(A) Continuous & not differentiable at x = 1

(B) Continuous & differentiable at x = 1

(C) Discontinuous & differentiable at x = 1

(D) Discontinuous & not differentiable at x = 1

Ans: B

5. If $f (x) = {x x^{2} + b x + c, x \leq 1, x > 1$

then find the values of b and c if f(x) is differentiable at x = 1

(A)b=-2, c=1 (B) b=-2, c=-2 (C) b=-1, c=2 (D)b=-1, c=1

Ans: D

7.0Continuity and Differentiability Solved Questions

1. What are the conditions for differentiability?

Ans: A function f(x) is differentiable at x = a if the limit $lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$ exists

2. What is continuity in a function?

Ans: Continuity means a function has no breaks, jumps, or holes in its graph. A function f(x) is continuous at x = a if $lim_{x \to a} f (x) = f (a)$ .

Frequently Asked Questions

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I authorise ALLEN Career Institute Pvt Ltd to send me regular updates via Phone calls, Whatsapp, SMS, Robocalls (Automated Calls), Emails, or on postal address.

Continuity and Discontinuity

Continuity in mathematics means a function has no interruptions at a point, meeting three conditions: defined at the point, limit exists, and limit equals the function's value. Discontinuity occurs when these...

Continuity

Continuity of a function is a fundamental concept in mathematics that describes the connectedness of the graph of the function. Formally, a...

Determinant of a Matrix

The determinant of a matrix is a single numerical value that provides important insights into the properties and behavior of the...

Determinants and Matrices

Determinants are special numbers calculated from square matrices. They help find areas,...

Differential Equations

An equation that involves independent and dependent variables and the derivatives of the dependent variables is called a Differential Equation...

Domain and Range of a Relation

The domain is all the possible input values, while the range is all the possible output values. Knowing how...

Inverse Trigonometric Functions

Inverse Trigonometric Functions are the inverse counterparts of the fundamental trigonometric functions: sine, cosine,...

Continuity and Differentiability

Continuity and differentiability are key concepts in calculus. Continuity means a function has no breaks in its graph. Differentiability means the function's slope can be defined at every point. A function must be continuous to be differentiable, but not vice versa. Understanding these concepts helps in analyzing functions and solving mathematical problems effectively.

1.0Continuity

Continuity of a function is a fundamental concept in mathematics that describes the smoothness of the graph of the function. Formally, a function f(x) is considered to be continuous at a point x = c if three conditions are met:

f(c) is defined, meaning that the function has a value at x = c.
The function has a defined limit as x approaches the value c. This means that as x gets arbitrarily close to c when the values of f(x) approach a specific finite value.
The value of the function at x = c coincides with the limit of the function at x = c, i.e.,

$limf_{x \to c} (x) = f (c)$ .

A function f is continuous at x = c if it is defined at x = c and the value of f(c) equals the limit of f(x) as x approaches c. If f is not continuous at c, then f is said to be discontinuous at c, and c is referred to as a point of discontinuity of f.

Also Read Continuity in Interval

2.0Differentiability

In Calculus, a function f(x) is considered differentiable at a point x = c if its derivative exists at that point. This entails the limit of the difference quotient $\frac{f ( x ) - f ( c )}{x - c}$ approaching a finite value as x approaches c. Graphically, differentiability implies a smooth, continuous curve with a well-defined tangent line at c, f(c).

The derivative of a function f at x = c is defined by the limit: $f^{'} (c) = lim_{h \to 0} \frac{f ( c + h ) - f ( c )}{h}$ provided this limit exists. The derivative of f at x = c is denoted by f'(c). The function defined by $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ wherever the limit exists, is called the derivative of f. The derivative of f is denoted by f'(x), $\frac{d}{dx} f (x)$ , or if y = f(x), by $\frac{dy}{dx} or y^{'}$ .The process of finding the derivative of a function is known as differentiation. We also say, "differentiate f(x) with respect to x " to mean finding f'(x).

3.0Algebra of Continuous Function

Suppose u and v be two real functions continuous at a real number c. Then

u + v is continuous at x = c.
u – v is continuous at x = c.
u.v is continuous at x = c.
$\frac{u}{v}$ is continuous at x = c, (provided v(c) ≠ 0).

Also Read Continuity and Discontinuity

4.0Algebra of Derivative Function

The following rules were established as a part of algebra of derivatives:

(u ± v) ' = u' ± v'
(u.v)' = u'v + uv' (Leibnitz or product rule)
$(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ , (Quotient rule)

5.0Continuity and Differentiability Solved Examples

Example 1: Discuss the continuity of $f (x) = ⎩ ⎨ ⎧ ∣ x + 1∣, 2 x + 3, x^{2} + 3, x^{3} - 15, x < - 2 - 2 \leq x < 0 0 \leq x < 3 x \geq 3$

Solution:

We write f(x) as $f (x) = ⎩ ⎨ ⎧ - x - 1, 2 x + 3, x^{2} + 3, x^{3} - 15, x < - 2 - 2 \leq x < 0 0 \leq x < 3 x \geq 3$

As we can see, f(x) is defined as a polynomial function in every interval (–∞, –2), [0, 3), [–2, 0), and [3, ∞). Consequently, it is continuous in each of these 4 open intervals. Thus, we check the continuity at

x = –2, 0, 3

At the point

$lim_{x \to - 2^{-}} f (x) = lim_{x \to - 2^{-}} (- x - 1) = + 2 - 1 = 1$

$lim_{x \to - 2^{+}} f (x) = lim_{x \to - 2^{+}} (2 x + 3) = 2 \cdot (- 2) + 3 = - 1$

Therefore, $lim_{x \to - 2} f (x)$ does not exist and hence f(x) is discontinuous at .

At the point

$lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} (2 x + 3) = 3$

$lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} (x^{2} + 3) = 3 f (0) = 0^{2} + 3 = 3$

Therefore f(x) is continuous at x = 0 .

At the point x = 3

$lim_{x \to 3^{-}} f (x) = lim_{x \to 3^{-}} (x^{2} + 3) = 3^{2} + 3 = 12$

$lim_{x \to 3^{+}} f (x) = lim_{x \to 3^{+}} (x^{3} - 15) = 3^{3} - 15 = 12$

f(3)=3^3-15=12

Therefore, f(x) is continuous at x = 3 .

We find that f(x) is continuous at all points in $R$ except at x = –2

Example 2: If $f (x) = {sin \frac{p x}{2}, [x] x < 1 x \geq 1$ then find whether f(x) is continuous or not at x = 1 , where [.] denotes greatest integer function.

Solution:

$f (x) = {sin \frac{p x}{2}, x < 1 [x], x \geq 1$

For continuity at x = 1, we determine, $f (1) lim_{x \to 1^{-}} f (x) and lim_{x \to 1^{+}} f (x)$ .

Now, f(1)=[1]=1 $f (x) = lim_{x \to 1^{-}} sin \frac{p x}{2} = sin \frac{p}{2} = 1 and lim_{x \to 1^{+}} f (x) = lim_{x \to 1^{+}} [x] = 1$

so $f (1) = lim_{x \to 1^{-}} f (x) = f (x)$

∴ f(x) is continuous at

Example 3:

If $f (x) = \frac{x + 1}{x - 1} and g (x) = \frac{1}{x - 2}$ , then discuss the continuity of f(x), g(x) and fog(x) in its domain.

Solution:

$f (x) = \frac{x + 1}{x - 1}$

f(x) is a rational function it must be continuous in its domain and f is not defined at .

$g (x) = \frac{1}{x - 2}$

g(x)is also a rational function. It must be continuous in its domain and g is not defined at x = 2 .

Consider g(x) = 1

$\frac{1}{x - 2} = 1 \Rightarrow x = 3$

∴ fog (x)is continuous in its domain: R – {2, 3}

Example 4: If f(x)= $[x - x if if x \in R x \in / R$ find the points where f(x) is continuous.

Solution:

Let be the point at which f(x) is continuous.

⇒ $lim_{x \to a throuthetraional} f (x) = lim_{x \to a throughitrational} f (x)$

⇒ a = –a

⇒ a = 0 ⇒ function is continuous at x = 0.

Example 5: Determine the values of x for which the following functions fails to be continuous or differentiable

$f (x) = ⎩ ⎨ ⎧ (1 - x), (1 - x) (2 - x), (3 - x), x < 1 1 \leq x \leq 2 x > 2 .$ Justify your answer.

Solution:

By the given definition the function f is continuous and differentiable at all points except possibility at x = 1 and x = 2.

Check the differentiability at x = 1

$\Rightarrow q = LHD = lim_{h \to 0} \frac{f ( 1 - h ) - f ( 1 )}{- h} = lim_{h \to 0} \frac{1 - ( 1 - h ) - 0}{- h} = - 1$

$\Rightarrow p = RHD = lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h} = lim_{h \to 0} \frac{{ 1 - ( 1 + h )} { 2 - ( 1 + h ) - 0 }}{h} = - 1$

⇒ $q = p or L . H . S = R . H . S$

∴ Differentiable at x =1 ⇒ Continuous at x = 1 .

Check the differentiability at x = 2

⇒ $q = LHD = lim_{h \to 0} \frac{f ( 2 - h ) - f ( 2 )}{- h} = lim_{h \to 0} \frac{( 1 - 2 + h ) ( 2 - 2 + h ) - 0}{- h} = 1 = finite$

⇒ $p = R HD = lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{( 3 - 2 - h ) - 0}{h} \to \infty (n o t f ini t e) ∵ q \neq = p$ ∴ not differentiable at x = 2 .

Now we have to check the continuity at

$\Rightarrow LHL = lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{-}} (1 - x) (2 - x) = lim_{h \to 0} (1 - (2 - h)) (2 - (2 - h)) = 0$

$\Rightarrow RHL = lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} (3 - x) = lim_{h \to 0} (3 - (2 + h)) = 1$

∴ LHL ≠ RHL

⇒ not continuous at x = 2 .

Example 6: f(x)=|| x|-1| is not differentiable at

(A) 0 (B) ±1,0 (C) 1 (D) ±1

Ans. (B)

Solution:

= ${∣ x ∣ - 1, - ∣ x ∣ + 1, ∣ x ∣ - 1 \geq 0 ∣ x ∣ - 1 < 0$

= ${∣ x ∣ - 1, - ∣ x ∣ + 1, x \leq - 1 or x \geq 1 - 1 < x < 1$

$= ⎩ ⎨ ⎧ - x - 1, x + 1, - x + 1, x - 1, x \leq - 1 - 1 < x < 0 0 \leq x < 1 x \geq 1$

From the graph. It is clear that f(x) is not differentiable at x = –1,0 and 1.

Example 7: For what triplets of real numbers (a,b,c) with a $\neq = 0$ the function

$f (x) = {x, a x^{2} + b x + c, x \leq 1 otherwise$

is differentiable for all real x ?

(A) ${(a, 1 - 2 a, a) ∣ a \in R, a \neq = 0} (B) {(a, 1 - 2 a, c) ∣ a, c \in R, a \neq = 0}$

(C) ${(a, b, c) ∣ a, b, c \in R, a + b + c = 1} (D) {(a, 1 - 2 a, 0) ∣ a \in R, a \neq = 0}$

Ans. (A)

Solution:

$f (x) = {x a x^{2} + b x + c,, x \leq 1 otherwise$

⇒ f(x) should be continuous at x=1

it gives a + b + c = 1

⇒ f(x)should be differentiable at x =1

it gives $2 a + b = 1 \Rightarrow b = 1 - 2 a$

⇒ $c = 1 - a - b = a$

Example 8: If $f (x) = ⎩ ⎨ ⎧ e^{- ∣ x ∣}, - e^{- ∣ x - 1∣} + e^{- 1} + 1, e^{- ∣ x - 2∣}, - 5 < x < 0 0 \leq x < 2 2 \leq x < 4$

Discuss the continuity and differentiability of f(x) in the interval (–5, 4).

Solution:

$f (x) = ⎩ ⎨ ⎧ e^{+ x} - e^{x - 1} + e^{- 1} + 1 - e^{- x + 1} + e^{- 1} + 1 e^{- x + 2} - 5 < x < 0 0 \leq x \leq 1 1 < x < 2 2 \leq x < 4$

Check the differentiability at x = 0

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 0 - h ) - f ( 0 )}{- h} = lim_{h \to 0} \frac{e ^{- h} - 1}{- h} = 1$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 0 + h ) - f ( 0 )}{h} = lim_{h \to 0} \frac{- e ^{h - 1} + e ^{- 1} + 1 - 1}{h} = - e^{- 1}$

$\Rightarrow LHD \neq = RHD$

∴ Not differentiable at , but continuous at since LHD and RHD both are finite.

Check the differentiability at x = 1

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 1 - h ) - f ( 1 )}{- h} = lim_{h \to 0} \frac{- e ^{1 - h - 1} + e ^{- 1} + 1 - e ^{- 1}}{- h} = - 1$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 1 + h ) - f ( 1 )}{h} = lim_{h \to 0} \frac{- e ^{1 - h - 1} + e ^{- 1} + 1 - e ^{- 1}}{h} = 1$

LHD ≠ RHD

∴ Not differentiable at x = 1 , but continuous at x = 1 since LHD and RHD both are finite.

Check the differentiability at x = 2

$\Rightarrow LHD = lim_{h \to 0} \frac{f ( 2 - h ) - f ( 2 )}{- h} = lim_{h \to 0} \frac{- e ^{- 2 + h + 1} + e ^{- 1} + 1 - 1}{- h} = lim_{h \to 0} \frac{- e ^{- 1} ( e ^{h} - 1 )}{- h} = e^{- 1}$

$\Rightarrow RHD = lim_{h \to 0} \frac{f ( 2 + h ) - f ( 2 )}{h} = lim_{h \to 0} \frac{e ^{- h} - 1}{h} = - 1$

LHD ≠ RHD

∴ Not differentiable at x = 2, but continuous at x = 2 since LHD & RHD both are finite.

Example 9: The function $f (x) = (x^{2} - 1) x^{2} - 3 x + 2 + cos (∣ x ∣)$ is NOT differentiable at:

(A) –1 (B) 0 (C) 1 (D) 2

Ans. (D)

Solution:

$f (x) = (x^{2} - 1) ∣ (x - 1) (x - 2) ∣$

Clearly not differentiable at x = 2

Also Read Differential Equation previous year questions with solutions

6.0Continuity and Differentiability Practice Problems

1. The function $f (x) = \frac{4 - x ^{2}}{4 x - x ^{3}}$ , is -

(A) discontinuous at only one point in its domain.

(B) discontinuous at two points in its domain.

(C) discontinuous at three points in its domain.

(D) continuous everywhere in its domain.

Ans: C

2. A function y = f(x) is defined as $f (x) = {k sin \frac{( x + 3 ) p}{6} \frac{3 - 11 - x}{x - 2} for for x \leq 2 x > 2$ . If f(x) is continuous at x = 2, then find the value of k .

Ans: $k = \frac{1}{3}$

3. A function is defined as follows

$f (x) = {x^{3}; x; x^{2} < 1 x^{2} \geq 1$

is

(A) Continuous but not differentiable at x = 1

(B) Discontinuous but differentiable at x = 1

(C) Discontinuous but not differentiable at x = 1

(D) Continuous but differentiable at x = 1

Ans: A

4. A function is defined as follows $f (x) = (x - 1) ∣ x - 1∣$ .

(A) Continuous & not differentiable at x = 1

(B) Continuous & differentiable at x = 1

(C) Discontinuous & differentiable at x = 1

(D) Discontinuous & not differentiable at x = 1

Ans: B

5. If $f (x) = {x x^{2} + b x + c, x \leq 1, x > 1$

then find the values of b and c if f(x) is differentiable at x = 1

(A)b=-2, c=1 (B) b=-2, c=-2 (C) b=-1, c=2 (D)b=-1, c=1

Ans: D

7.0Continuity and Differentiability Solved Questions

1. What are the conditions for differentiability?

Ans: A function f(x) is differentiable at x = a if the limit $lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}$ exists

2. What is continuity in a function?

Ans: Continuity means a function has no breaks, jumps, or holes in its graph. A function f(x) is continuous at x = a if $lim_{x \to a} f (x) = f (a)$ .

Frequently Asked Questions

What are the types of discontinuities?

What is the significance of continuity in calculus?

What is the differentiability in a function?

How is differentiability related to continuity?

What is the geometric interpretation of differentiability?

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Table of Contents

Frequently Asked Questions

What are the types of discontinuities?

What is the significance of continuity in calculus?

What is the differentiability in a function?

How is differentiability related to continuity?

What is the geometric interpretation of differentiability?

Join ALLEN!

Related Articles:-

Continuity and Discontinuity

Continuity

Determinant of a Matrix

Determinants and Matrices

Differential Equations

Domain and Range of a Relation

Inverse Trigonometric Functions

Continuity and Differentiability

1.0Continuity

2.0Differentiability

3.0Algebra of Continuous Function

4.0Algebra of Derivative Function

5.0Continuity and Differentiability Solved Examples

6.0Continuity and Differentiability Practice Problems

7.0Continuity and Differentiability Solved Questions

Table of Contents