## Continuity on an Open Interval

Say a function be f(x) is said to be continuous on an open interval (a,b) iff it is continuous at all possible points on the interval (a,b).

### Continuity on a Closed Interval

On an open interval [a,b], a function f(x) is said to be continuous, iff

(1) f is continuous on the open interval of (a, b)

(2) \[\lim_{x\rightarrow a+}\] f(x) = f(a)

and (3) \[\lim_{x\rightarrow b-}\] f(x) = f(a)

In other words, f(x) is continuous on [a, b] iff it is continuous on (a, b) and it is continuous at a from the right and at b from the left.

Continuous functions:- A function f(x) is said to be continuous, if it is continuous at each point of its given domain.

Everywhere Continuous function: - A function f(x) is said to be everywhere continuous if it is continuous on entire real line (-∞,∞ ).

## Properties of Continuous Function

There are several theorems related to the continuity of a function in a given interval, which are as follows:

### Theorem 1:

If f and g are two continuous functions on their common domain D, then

(i) f + g is continuous on D

(ii) f - g is continuous on D

(iii) f g is continuous on D

(iv) α f is continuous on D, where α is any scalar.

(v) - is continuous on D – {x : g(x) ≠ 0}

(vi) ⊥ is continuous on D – {x : f(x) ≠ 0}

### Theorem 2:

The composition of two continuous functions is also a continuous function.

### Theorem 3:

If f is continuous in its given domain D, then I f I is also continuous on D.

### Theorem 4:

A constant function is everywhere continuous.

### Theorem 5:

The identity function is everywhere continuous.

### Theorem 6:

A polynomial function is continuous everywhere.

### Theorem 7:

The modulus function is continuous everywhere.

### Theorem 8:

The exponential function a\[^{x}\] , a > 0 is continuous everywhere.

### Theorem 9:

The logarithmic functions f(x) = log x are continuous in its given domain.

### Theorem 10:

The sine function y = sin x is continuous everywhere.

### Theorem 11:

The cosine function y = cos x is continuous everywhere.

### Theorem12:

The tangent function y = tan x is continuous everywhere.

### Theorem 13:

The cosecant function is everywhere in its given domain.

## Mean Value Theorems

These are the theorems related to continuity and differentiability of a function.

### Theorem:- Rolle’S Theorems

Let f(x) be any function of x defined on the closed interval [a,b] such that

On the closed interval [a,b] is a continuous function

On the open interval (a,b), f(x) is a differentiable function

such that

f(a) = f(b)

Then, there includes at least one point c in the open interval (a,b) such that f’(c)=0.

### Theorem 2:- Lagrange’s’ Mean Value Theorem

Let f(x) be any real function defined on the closed interval [a,b] such that

The function is continuous on all the possible points in the closed interval [a,b]

The function is differentiable on all possible points the open interval (a,b)

Then, we can find a real number c ∈ (a, b) such that

f’(c) = \[\frac{f(b) - f(a)}{b - a}\]

## Solved Examples on Continuity -In-Interval

### 1. Evaluate the Continuity of the Given Polynomial Function

f(x) 2x - 1, x < 0

2x + 1, x ≥ 0

### Solution:

Given here, when x <0 the function is f(x) = 2x-1 .

Now, we can see here that f(x) is a polynomial function of degree 1 which is continuous at all x < 0.

When x > 0 the function is

f(x)= 2x+1 which is again a polynomial function and hence continuous at all

x > 0. Now let’s take x = 0 , we see that

\[\lim_{x\rightarrow 0-}\] f(x) = \[\lim_{x\rightarrow 0}\] (2x - 1) = -1

\[\lim_{x\rightarrow 0+}\] f(x) = \[\lim_{x\rightarrow 0}\] (2x + 1) = 1

Clearly, we can see that L.H.L≠R.H.L,

\[\lim_{x\rightarrow 0-}\] f(x) ≠ \[\lim_{x\rightarrow 0+}\] f(x)

Therefore, at x = 0

f(x) is not continuous

### 2. Discuss the Point of Discontinuity for the Given Function.

f(x) = \[\frac{sin 3x}{x}\] , if x ≠ 0

and 4, if x = 0

### Solution:

We can see that the function is a sine function which is continuous everywhere in its domain.

Now we need to find the L.H.L and R.H.L of the given function

R.H.L = \[\lim_{x\rightarrow 0-}\] f(x) = \[\lim_{x\rightarrow 0}\] \[\frac{sin 3x}{x}\] x \[\frac{3}{3}\] = 3\[\lim_{x\rightarrow 0}\] \[\frac{sin 3x}{x}\] = 3

L.H.L = \[\lim_{x\rightarrow 0+}\] f(x) = \[\lim_{x\rightarrow 0}\] \[\frac{sin 3x}{x}\] x \[\frac{3}{3}\] x \[\frac{3}{3}\] = 3 \[\lim_{x\rightarrow 0}\] \[\frac{sin 3x}{x}\] = 3

f(0) = 4

Clearly we can see that L.H.L ≠ R.H.L ≠ f(0) at x = 0.

Hence function f(x) is discontinuous at x=0.

### Did You Know

Any differentiable function must be a continuous function but any continuous variable is not always true.

There are rules to solve the indeterminate forms of limits such as L'hospital rule which includes the differentiation of numerator as well as denominator till indeterminate form is not removed.

f’(c) = \[\frac{f(b) - f(a)}{b - a}\] will give you the slope of the chord joining two points.

Interpretation of Lagrange’s theorem

Let f(x) be a function defined on [a,b] such that it represents the equation of curve y=f(x) and it is a continuous curve between the points P(a, f(a)) and Q(b, f(b)) then there exists a unique tangent which parallels to the line joining the two points.

Interpretation of Rolle’s theorem

Let (x) be a polynomial function with a and b as its roots. We know that a polynomial function is everywhere continuous and differentiable and let a and b are roots of f(x), then f(a)=f(b)=0. So, we can say that f(x) satisfies the conditions of Rolle's theorem. Consequently , there exists a c ∊ (a,b) such that f’(c) = 0 i.e f’(x) = 0 at x=c.

Thus, Rolle’s can be interpreted as follows:

“We can find a number between any two roots of a polynomial that exits a root of its derivative f’(x).”

## FAQs on Continuity in Interval

Q1. When the Given Function is Continuous, Find the Value of k.

f(x) = kx+5, if x ≤ 2

and x-1 , if x>2

Solution: Here we can see that the given function is a polynomial function, which is a continuous function for all values of x.

So we will equate the limits and value of the function at x=2

lim_{x⟶2-} f(x) = lim_{x⟶2 }kx + 5 = 2k + 5

lim_{x⟶2+ }f(x) = lim_{x⟶2} x - 1 = 1

f(2)= kx+5= 2k+5

Since function is continuous then

L.H.L=R.H.L=f(2)

therefore , 2k+5=1

k=2

The value of k=2.

Q2. Find the Point of Discontinuity of the Given Function:-

f(x)= { sinx/x + cos x if x≠0

5 , if x=0

Solution: Given, the functions are trigonometric functions which are continuous for all x

Now to find the point of discontinuity we will have to find the limits

lim_{x⟶0-} f(x) = lim_{x⟶0} sinx/x + cos x

L.H.L = lim_{x⟶0} sin x/x + lim_{x⟶0} cos x = 1 + 1 = 2

lim_{x⟶0+} f(x) = lim_{x⟶0} sin x/x + cos x

R.H.L = lim_{x⟶0} sin x/x + lim_{x⟶0} cos x = 1 + 1 = 2

And f(0)=2

We can see that L.H.L≠R.H.L≠f(0)

So, it is clear that the point of discontinuity is 0.