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:

1. f(c) is defined, meaning that the function has a value at x = c.
2. 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.
3. The value of the function at x = c coincides with the limit of the function at x = c, i.e.,

${\mathrm{limf}}_{\mathrm{x}\to \mathrm{c}}(\mathrm{x})=\mathrm{f}(\mathrm{c})$ .

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.

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 $\frac{f(\mathrm{x})-\mathrm{f}(\mathrm{c})}{\mathrm{x}-\mathrm{c}}$  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: $f^{\prime }(c)={\lim }_{h\to 0}\frac{f(c+h)-f(c)}{h}$ provided this limit exists. The derivative of f at x = c is denoted by f'(c). The function defined by $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$ wherever the limit exists, is called the derivative of f. The derivative of f is denoted by f'(x), $\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{f}(\mathrm{x})$ , or if y = f(x), by $\frac{\mathrm{dy}}{\mathrm{dx}}\text{ or }{\mathrm{y}}^{\prime }$ .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

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

4.0Algebra of Derivative Function

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

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

5.0Continuity and Differentiability Solved Examples

Example 1: Discuss the continuity of $f(x)=\{\begin{array}{ll}|x+1|,& x<-2\\ 2x+3,& -2\le x<0\\ x^2+3,& 0\le x<3\\ x^3-15,& x\ge 3\end{array}$ 

Solution:

We write f(x) as $f(x)=\{\begin{array}{ll}-x-1,& x<-2\\ 2x+3,& -2\le x<0\\ x^2+3,& 0\le x<3\\ x^3-15,& x\ge 3\end{array}$

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

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

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

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

At the point

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

${\lim }_{x\to 0^{+}}f(x)={\lim }_{x\to 0^{+}}\left(x^2+3\right)=3f(0)=0^2+3=3$

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

At the point x = 3 

${\lim }_{x\to 3^{-}}f(x)={\lim }_{x\to 3^{-}}\left(x^2+3\right)=3^2+3=12$

${\lim }_{x\to 3^{+}}f(x)={\lim }_{x\to 3^{+}}\left(x^3-15\right)=3^3-15=12$

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 $\mathbb{R}$ except at x = –2 

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

Solution:

$f(x)=\{\begin{array}{l}\sin \frac{\mathrm{p}x}{2},x<1\\ [x],x\ge 1\end{array}$

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

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

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

∴  f(x) is continuous at

Example 3:

If $f(x)=\frac{x+1}{x-1}\text{ and }g(x)=\frac{1}{x-2}$ , 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)=$[\begin{array}{lll}x& \text{ if }& x\in \mathbb{R}\\ -x& \text{ if }& x\notin \mathbb{R}\end{array}$  find the points where f(x) is continuous.

Solution:

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

⇒${\lim }_{\begin{array}{c}x\to a\\ \text{ throuthetraional }\end{array}}f(x)={\lim }_{\begin{array}{c}x\to a\\ \text{ throughitrational }\end{array}}f(x)$

⇒  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

$f(x)=\{\begin{array}{ll}(1-x),& x<1\\ (1-x)(2-x),& 1\le x\le 2\\ (3-x),& x>2\end{array}.$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 

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

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

⇒ $q=porL.H.S=R.H.S$

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

Check the differentiability at x = 2 

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

⇒  $p=RHD={\lim }_{h\to 0}\frac{f(2+h)-f(2)}{h}={\lim }_{h\to 0}\frac{(3-2-h)-0}{h}\to \infty (notfinite)\because q\ne p$ ∴ not differentiable at x = 2 .

Now we have to check the continuity at  

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

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

∴ 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:

=$\{\begin{array}{ll}|x|-1,& |x|-1\ge 0\\ -|x|+1,& |x|-1<0\end{array}$

=$\{\begin{array}{ll}|x|-1,& x\le -1\text{ or }x\ge 1\\ -|x|+1,& -1<x<1\end{array}$

$=\{\begin{array}{ll}-x-1,& x\le -1\\ x+1,& -1<x<0\\ -x+1,& 0\le x<1\\ x-1,& x\ge 1\end{array}$

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 $\ne 0$  the function 

$f(x)=\{\begin{array}{cl}x,& x\le 1\\ a{x}^2+bx+c,& \text{ otherwise }\end{array}$

is differentiable for all real x ?

(A) $\{(a,1-2a,a)\mid a\in R,a\ne 0\}(B)\{(a,1-2a,c)\mid a,c\in R,a\ne 0\}$

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

Ans. (A)

Solution:

$f(x)=\{\begin{array}{ccc}x& ,& x\le 1\\ a{x}^2+bx+c& ,& \text{ otherwise }\end{array}$

⇒ f(x) should be continuous at x=1 

it gives a + b + c = 1 

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

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

⇒ $c=1-a-b=a$

Example 8: If $f(x)=\{\begin{array}{lc}{e}^{-|x|},& -5<x<0\\ -e^{-|x-1|}+e^{-1}+1,& 0\le x<2\\ e^{-|x-2|},& 2\le x<4\end{array}$

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

Solution:

$f(x)=\{\begin{array}{cc}{e}^{+x}& -5<x<0\\ -e^{x-1}+e^{-1}+1& 0\le x\le 1\\ -e^{-x+1}+e^{-1}+1& 1<x<2\\ e^{-x+2}& 2\le x<4\end{array}$

Check the differentiability at x = 0 

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

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

$\Rightarrow \mathrm{LHD}\ne \mathrm{RHD}$

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

Check the differentiability at x = 1 

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

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

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 

$\Rightarrow \mathrm{LHD}={\lim }_{h\to 0}\frac{f(2-h)-f(2)}{-h}={\lim }_{h\to 0}\frac{-e^{-2+h+1}+e^{-1}+1-1}{-h}={\lim }_{h\to 0}\frac{-e^{-1}\left(e^h-1\right)}{-h}=e^{-1}$

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

LHD ≠ RHD

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

Example 9: The function $f(x)=\left(x^2-1\right)\left|x^2-3x+2\right|+\cos (|x|)$ is NOT differentiable at:

(A) –1 (B) 0 (C) 1 (D) 2

Ans. (D)

Solution:

$f(x)=\left(x^2-1\right)|(x-1)(x-2)|$

Clearly not differentiable at x = 2 

6.0Continuity and Differentiability Practice Problems

1. The function $f(x)=\frac{4-x^2}{4x-x^3}$ , 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 $f(x)=\{\begin{array}{lll}k\sin \frac{(x+3)p}{6}& \text{ for }& x\le 2\\ \frac{3-\sqrt{11-x}}{x-2}& \text{ for }& x>2\end{array}$ . 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  

$f(x)=\{\begin{array}{ll}{x}^3;& x^2<1\\ x;& x^2\ge 1\end{array}$

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 $f(x)=\{\begin{array}{cl}x& ,x\le 1\\ x^2+bx+c& ,x>1\end{array}$

  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 ${\lim }_{x\to a}f(x)=f(a)$.