Relation between continuity and differentiability: Continuity refers to a definition of the concept of a function that varies without any breaks or jumps. If the derivative of a function exists at each point in its domain, it is said to be differentiable. This means that a function can be differentiated at any point where its derivative is defined.

There are many real-life applications of continuity and differentiability, such as when launching rockets, a force is necessary to push the rocket higher, and this force is known as thrust. The derivative of momentum is used to calculate thrust. Doctors employ derivatives to determine the rate of tumour growth. This article teaches us about the relationship between continuity and differentiability.

Introduction to Continuity

Consider the graph of the function \(y=f(x)\) as given below.

It is evident from the graph that:

i. \(f(x)\) is not defined at \(x=a.\)

ii. Since \(L = R \Rightarrow \mathop {\lim }\limits_{x \to {a^ – }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)\) i.e. Left hand limit= Right hand limit at \(x=a\) and so \(\mathop {\lim }\limits_{x \to {a^ – }} f(x)\) exists.

iii. This function is not continuous because \(f(a)\) is not defined.

Consider the function whose graph is shown below.

It is evident from the above graph that:

i. The value of the function at \(x=a\) is \(f(a)\)

ii. \(L = R \ne f(a)i.e.,\mathop {\lim }\limits_{x \to {a^ – }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) \ne f(a)\)

iii. This function is not continuous because \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) \ne f(a)\)

Now, consider the graph is given below.

It is evident from the above graph that:

i. \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) \ne \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\)

ii. The function is not continuous because \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) \ne \mathop {\lim }\limits_{x \to {a^ + }} f(x)\)

The continuity of a function \(f\) is also destroyed if either of the two limits \(\mathop {\lim }\limits_{x \to {a^ – }} f(x)\) and \(\mathop {\lim }\limits_{x \to {a^ + }} f(x)\) both tend to \( + \infty \) or \( – \infty \) and \(f(\alpha)\) is finite.

Continuity at a Point

A function \(f(x)\) is said to be continuous at the point \(x=a\) if \(\mathop {\lim }\limits_{x \to {a }} f(x)=f(a)\)

Thus, continuity of \(f(x)\) at \(x=a\) implies that:

i. \(f(x)\) is defined at \(x=a\) i.e., \(a\) is in the domain of \(f\).

ii. \(\mathop {\lim }\limits_{x \to {a }} f(x)\) exists i.e., \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x)\)

iii. \(\mathop {\lim }\limits_{x \to a} f(x) = f(a)\)

Discontinuity at a Point

A function \(f\) fails to be continuous at \(x=a\) (i.e., is discontinuous at \(x=a\) ) for any of the following reasons:

i. \(f\) is not defined at \(x=a, a \notin \mathrm{D}_{f}\)

ii. \(\mathop {\lim }\limits_{x \to {a }} f(x)\) does not exist i.e., \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) \ne \mathop {\lim }\limits_{x \to {a^ + }} f(x)\)

iii. \(\mathop {\lim }\limits_{x \to {a }} f(x)\) exists but \(\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\)

Continuous Function

A function \(f\) is said to be a continuous function if it is continuous at every point of its domain.

Domain of Continuity

The set of all points where a function \(f\) is continuous is called the domain of continuity of \(f\).

Left Continuity at a Point

The function \(f\) is said to be left continuous at \(x=a\) if \(\mathop {\lim }\limits_{x \to {a^ – }} f(x)=f(a)\)

Right Continuity at a Point

The function \(f\) is said to be right continuous at \(x=a\) if \(\mathop {\lim }\limits_{x \to {a^ + }} f(x)=f(a)\)

Note: \(f\) is continuous at \(x=a\), if \(f\) is left continuous and right continuous at \(x=a\)

Continuity in an Open Interval

A function \(f\) is said to be continuous in the open interval \((a,b)\) if \(f\) is continuous at every point in the interval \((a,b)\).

Continuity in a Closed Interval

A function \(f\) is said to be continuous in the closed interval \([a,b]\) if

i. \(f\) is continuous in the open interval \((a,b)\)

ii. \(f\) is right continuous at left endpoint \(a\), i.e. \(\mathop {\lim }\limits_{x \to {a^ +}} f(x)=f(a)\)

iii. \(f\) is left continuous at right endpoint \(b\), i.e. \(\mathop {\lim }\limits_{x \to {b^ – }} f(x)=f(b)\)

Removable Discontinuity

A removable discontinuity occurs at a point at which the limit of the function exists, i.e. the left-hand limit of the function at some point equals the right-hand limit of the function at that point, but it does not equal the value of the function at that point.

Consider the piece-wise function \(f(x) = \left\{ {\begin{array}{*{20}{c}}{{x^2}}&{for}&{x < 1}\\0&{for}&{x = 1}\\{2 – x}&{for}&{x > 1}\end{array}} \right\}\)

The graph of this function is shown below.

From the graph, it is evident that \(\mathop {\lim }\limits_{x \to x^-_0 } f(x) = \mathop {\lim }\limits_{x \to x_0^ + } f(x) \ne f\left( {{x_0}} \right)\)

Thus, this function is not continuous at \(x=x_{0}\) and this is an example of removable discontinuity.

Let \(\mathop {\lim }\limits_{x \to x^-_0 } f(x) = \mathop {\lim }\limits_{x \to x_0^ + } f(x) = L\) then we can remove the continuity of the function by defining it as follows: \(g(x) = \left\{ {\begin{array}{*{20}{c}}{f(x)}&{x \ne {x_0}}\\L&{x = {x_0}}\end{array}} \right\}\)

Thus, we can say that if \(x=a\) is a point of discontinuity of \(f\) such that \(\mathop {\lim }\limits_{x \to {a}} f(x)\) exist but \(\mathop {\lim }\limits_{x \to a} f(x) \ne f(a)\), then \(f\) is said to have a removable discontinuity at \(x=a\) This is because \(f\) can be made continuous at \(x=a\) by redefining \(f\) at a such that \(f(a)=\mathop {\lim }\limits_{x \to {a}}f(x)\)

Steps to Solve Removable Discontinuity

Step 1: Factor out the numerator and the denominator.

Step 2: Determine the common factors in the numerator and the denominator.

Step 3: Set the common factors equal to zero.

Step 4: Solve for \(x\).

Step 5: The value of \(x\) is the required point of removable discontinuity.

Differentiability at a Point

Let \(f(x)\) be a real-valued function defined on an open interval \((a,b)\) and let \(c \in(a, b)\) Then \(f(x)\) is said to be differentiable at \(x=c \Longleftrightarrow \mathop {\lim }\limits_{x \to {c}} \frac{f(x)-f(c)}{x-c}\)exists finitely.

This limit is called the derivative or differential coefficient of the function \(f(x)\) at \(x=c\) and is denoted by \(f'(c)\) or \(\left(\frac{d f(x)}{d x}\right)_{x=c}\)

Thus, \(f^{\prime}(c)=\mathop {\lim }\limits_{x \to {c}}\frac{f(x)-f(c)}{x-c}\)

Now, \(f(x)\) is differentaible at \(x=c\)

\(\Rightarrow \mathop {\lim }\limits_{x \to {c }} \frac{f(x)-f(c)}{x-c}\) exists finitely

\( \Rightarrow \mathop {\lim }\limits_{x \to {c^ – }} \frac{{f(x) – f(c)}}{{x – c}} = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{f(x) – f(c)}}{{x – c}}\)

\(\Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{f(c – h) – f(c)}}{{ – h}} = \mathop {\lim }\limits_{h \to 0} \frac{{f(c + h) – f(c)}}{h}\)

Here,

• \(\mathop {\lim }\limits_{x \to {c^ – }}\frac{f(x)-f(c)}{x-c}\) or \(\mathop {\lim }\limits_{h \to {0}}\frac{f(c-h)-f(c)}{-h}\) is called the left-hand derivative of \(f(x)\) at \(x=c\) and is denoted by \(f^{\prime}\left(c^{-}\right)\) or \(L f^{\prime}(c)\).
• \(\mathop {\lim }\limits_{x \to {c^ + }}\frac{f(x)-f(c)}{x-c}\) or \(\mathop {\lim }\limits_{h \to {0}}\frac{f(c+h)-f(c)}{h}\) is called the right-hand derivative of \(f(x)\) and is denoted by \(f^{\prime}\left(c^{+}\right)\) or \(R f^{\prime}(c)\)

Thus,

\(f(x)\) is differentiable at \(x=c \Leftrightarrow L f^{\prime}(c)=R f^{\prime}(c)\).

Continuity vs Differentiability

1. For a function to be continuous \(\mathop {\lim }\limits_{x \to {a }} f(x)\) and \(\mathop {\lim }\limits_{x \to {a }} f(x)=f(a)\) for all points \(a\).

2. A function is differentiable anywhere its derivative is defined. A function \(f(x)\) is said to be differentiable at \(x=c \Longleftrightarrow \mathop {\lim }\limits_{x \to {c }} \frac{f(x)-f(c)}{x-c}\) exists finitely.

3. Differentiability is when the slope of the tangent line equals the limit of the function at a given point.

4. For a function to be differentiable, it must be continuous, and its derivative must be continuous.

5. Continuous graphs may contain cusps or vertical segments with undefined slopes,.

6. Not all continuous functions are differentiable.

Theorem of Continuity and Differentiability

Statement: If a function is differentiable at a point, it is necessarily continuous at that point. But the converse is not necessarily true.

Or,

\(f(x)\) is differentiable at \(x=c\), then \(f(x)\) is continuous at \(x=c\).

Proof: Let the function \(f(x)\) be differentiable at \(x=c.\)

\(\Rightarrow \mathop {\lim }\limits_{x \to {c }} \frac{f(x)-f(c)}{x-c}\) exists finitely.

Let \(\mathop {\lim }\limits_{x \to {c }} \frac{f(x)-f(c)}{x-c}=f^{\prime}(c)\)

To prove that \(f(x)\) is continuous at \(x=c\), it is sufficient to show that \(\mathop {\lim }\limits_{x \to {c }} f(x)=f(c)\)

Now,

\(\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \left\{ {\left( {\frac{{f(x) – f(c)}}{{x – c}}} \right)(x – c) + f(c)} \right\}\)

\( = \mathop {\lim }\limits_{x \to c} \left\{ {\left( {\frac{{f(x) – f(c)}}{{x – c}}} \right)(x – c)} \right\} + f(c)\)

\( = \mathop {\lim }\limits_{x \to c} \left( {\frac{{f(x) – f(c)}}{{x – c}}} \right) \times \mathop {\lim }\limits_{x \to c} (x – c) + f(c)\)

\(=f^{\prime}(c) \times 0+f(c)\)

\(\therefore \mathop {\lim }\limits_{x \to {c }} f(x)=f(c)\)

Hence, proved that \(f(x)\) is continuous at \(x=c\)

Converse of Theorem

Statement: The converse of this theorem is not necessarily true, i.e. a function may be continuous at a point but may not be differentiable at that point.

Example: The function \(f(x)=|x|\) is continuous at \(x=0\), but not differentiable at \(x=0\)

1. Check for continuity:

Now, \(f(x) = |x| = \left\{ {\begin{array}{*{20}{c}}x&{if}\\{ – x}&{if}\end{array}\begin{array}{*{20}{c}}{x \ge 0}\\{x < 0}\end{array}} \right\}\)

Left-hand limit \(=\mathop {\lim }\limits_{x \to {0^- }} f(x)\)

\(=\mathop {\lim }\limits_{h \to {0 }} f(0-h)\)

\(=\mathop {\lim }\limits_{h \to {0 }} [-(0-h)]\)

\(=\mathop {\lim }\limits_{h \to {0 }} h\)

\(\therefore\) L. H. L. \(=0\)

Now,

Right-Hand limit \(=\mathop {\lim }\limits_{x \to {0^+ }} f(x)\)

\( = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right)\)

\(\therefore \mathrm{R} . \mathrm{H} . \mathrm{L} .=0\)

We know that, \(f(0)=0\)

Since, \(\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)\)

Hence, the function is continuous at \(x=0\)

Now,

2. Check for Differentiability:

(L.H.D at \(x = 0) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{f(x) – f(0)}}{{x – 0}}\)

\(\Rightarrow(\) L. H. D. at \(x = 0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 – h) – f(0)}}{{0 – h – 0}}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{f( – h) – f(0)}}{{ – h}}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{| – h|}}{{ – h}}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{h}{{ – h}}\)

\(\therefore(\) L. H. D. at \(x=0)=-1\)

Thus, L.H.D. at \(x=0\) is \(-1\).

And,

(R. H. D.at \(x = 0) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) – f(0)}}{{x – 0}}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) – f(0)}}{h}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( h \right) – f\left( 0 \right)}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{|h| – |0|}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{h}{h}\)

\(\therefore(\) R. H. D. at \(x=0)=1\)

Thus, R.H.D at \(x=0\) is \(1\)

Since, (L. H. D.at \(x=0) \neq(\) R. H. D.at \(x=0)\)

Therefore, \(f(x)\) is not differentiable at \(x=0\)

Conclusion: Thus,all differentiable functions are continuous, but not all continuous functions are differentiable.

Solved Examples – Relation Between Continuity and Differentiability

Q.1. Show that \(f(x)=x^{2}\) is differentiable at \(x=1\) and find \(f^{\prime}(1)\)
Sol: Given \(f(x)=x^{2}\)
(L.H.D. at \(x = 1) = \mathop {\lim }\limits_{x \to {1^ – }} \frac{{f(x) – f(1)}}{{x – 1}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 – h) – f(1)}}{{1 – h – 1}}\)
\(= \mathop {\lim }\limits_{h \to 0} \frac{{{{(1 – h)}^2} – {1^2}}}{{ – h}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{ – 2h + {h^2}}}{{ – h}}\)
\( = \mathop {\lim }\limits_{h \to 0} (2 – h) = 2\)
Thus, L.H.D. at \(x=1\) is \(2\)
Since, (R.H.D. at \(x = 1) = \mathop {\lim }\limits_{x \to {1^ + }} \frac{{f(x) – f(1)}}{{x – 1}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h) – f(1)}}{{1 + h – 1}}\)
\(= \mathop {\lim }\limits_{h \to 0} \frac{{{{(1 + h)}^2} – 1}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{2h + {h^2}}}{h}\)
\( = \mathop {\lim }\limits_{h \to 0} (2 + h)\)
\(=2\)
Thus, R.H.D. at \(x=1\) is \(2\).
Since (L.H.D at \(x=1)=\) R.H.D at \(x=1)=2\)
Therefore, \(f(x)\) is differentiable at \(x=1\) and \(f^{\prime}(1)=2\)

Q.2. Show that the function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} \begin{gathered} x\sin \frac{1}{x} \hfill \\ 0 \hfill \\ \end{gathered} &\begin{gathered} {\text{when}} \hfill \\ {\text{when}} \hfill \\ \end{gathered} &\begin{gathered} x \ne 0 \hfill \\ x = 0 \hfill \\ \end{gathered} \end{array}} \right.\) is continuous but not differentiable at \(x=0\)
Sol:
1. Check for Continuity:
\(\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 – h)\)
\( = \mathop {\lim }\limits_{h \to 0} \left[ {(0 – h)\sin \frac{1}{{(0 – h)}}} \right]\)
\( = \mathop {\lim }\limits_{h \to 0} \left[ { – h\sin \left( {\frac{{ – 1}}{h}} \right)} \right]\)
\( = \mathop {\lim }\limits_{h \to 0} \left[ {h\sin \frac{1}{h}} \right]\)
\(=0 \times(\)an oscillating number between \(-1\) and \(1)\)
\(=0\)
\(\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h)\)
\( = \mathop {\lim }\limits_{h \to 0} f(h)\)
\( = \mathop {\lim }\limits_{h \to 0} h\sin \left( {\frac{1}{h}} \right)\)
\(=0 \times(\) an oscillating number between \(-1\) and \(1)\)
\(=0\)
Thus, we obtain \(\mathop {\lim }\limits_{x \to {0^ – }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)\)
Therefore, \(f(x)\) is continuous at \(x=0\)
2. Check for Differentiability:
(L.H.D. at \(x = 0) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{f(x) – f(0)}}{{x – 0}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 – h) – f(0)}}{{0 – h – 0}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{f( – h) – 0}}{{ – h}}\)
\( = \mathop {\lim }\limits_{h \to 0} \frac{{ – h\sin \left( {\frac{1}{{ – h}}} \right)}}{{ – h}}\)
\( = – \mathop {\lim }\limits_{h \to 0} \sin \left( {\frac{1}{h}} \right)\)
\(=\) A number which oscillates between \(-1\) and \(1\)
\(\therefore(\) L.H.D. at \(x=0)\) does not exist.
Similarly, it can be shown that R.H.D. at \(x=0\) does not exist.
Hence, \(f(x)\) is not differentiable at \(x=0\).

Q.3. Discuss the differentiability of \(f(x)=x|x|\) at \(x=0\)
Sol: Given \(f(x)=x|x|\)
\(f\left( x \right) = x\left| x \right| = \left\{ {\begin{array}{*{20}{c}} {{x^2}}&{x \geqslant 0} \\ { – {x^2}}&{x < 0} \end{array}} \right.\)
Now, (L.H.D. at \(x = 0) = \mathop {\lim }\limits_{x \to {0^ – }} \frac{{f(x) – f(0)}}{{x – 0}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{ – {x^2} – 0}}{{x – 0}}\)
\( = \mathop {\lim }\limits_{x \to 0} ( – x)\)
\(=0\)
And, (R.H.D. at \(x = 0) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) – f(0)}}{{x – 0}}\)
\( = \mathop {\lim }\limits_{x \to 0} \frac{{{x^2} – 0}}{{x – 0}}\)
\( = \mathop {\lim }\limits_{x \to 0} x\)
\(=0\)
Since (L.H.D. at \(x=0)=(\) R.H.D. at \(x=0)\)
Hence, \(f(x)\) is differentiable at \(x=0\).

Q.4. If \(f(x)\) is differentiable at \(x=a\) find \(\mathop {\lim }\limits_{x \to a} \frac{{{x^2}f\left( a \right) – {a^2}f\left( x \right)}}{{x – a}}\)
Sol: It is given that \(f(x)\) is differentiable at \(x=a\). Therefore, \(\mathop {\lim }\limits_{x \to a} \frac{{f(x) – f(a)}}{{x – a}}\) exists finitely.
Let \(\mathop {\lim }\limits_{x \to a} \frac{{f(x) – f(a)}}{{x – a}} = {f^\prime }(a)\)…..(i)
Now,\(\mathop {\lim }\limits_{x \to a} \frac{{{x^2}f(a) – {a^2}f(x)}}{{x – a}} = \mathop {\lim }\limits_{x \to a} \frac{{{x^2}f(a) – {a^2}f(a) + {a^2}f(a) – {a^2}f(x)}}{{x – a}}\)
\( = \mathop {\lim }\limits_{x \to a} \frac{{\left( {{x^2} – {a^2}} \right)f\left( a \right) – {a^2}\left\{ {f\left( x \right) – f\left( a \right)} \right\}}}{{x – a}}\)
\( = \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{\left( {{x^2} – {a^2}} \right)f(a)}}{{x – a}} – {a^2}\left( {\frac{{f(x) – f(a)}}{{x – a}}} \right)} \right\}\)
\( = \mathop {\lim }\limits_{x \to a} (x + a)f(a) – {a^2}\mathop {\lim }\limits_{x \to a} \frac{{f(x) – f(a)}}{{x – a}}\)
\(=2 a f(a)-a^{2} f^{\prime}(a)\) [Using (i)]
Hence, \(\mathop {\lim }\limits_{x \to a} \frac{{{x^2}f(a) – {a^2}f(x)}}{{x – a}} = 2af(a) – {a^2}{f^\prime }(a)\)

Q.5. For what choice of \(a\) and \(b\) is the function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^2}}&{x \leqslant c} \\ {ax + b}&{x > c} \end{array}} \right.\) is differentiable at \(x=c\)
Sol: It is given that \(f(x)\) is differentiable at \(x=c\) and every differentiable function is continuous. So, \(f(x)\) is continuous at \(x=c\).
\(\therefore \mathop {\lim }\limits_{x \to {c^ – }} f(x) = \mathop {\lim }\limits_{x \to {c^ + }} f(x) = f(c)\)
\( \Rightarrow \mathop {\lim }\limits_{x \to c} {x^2} = \mathop {\lim }\limits_{x \to c} (ax + b) = {c^2}\)
\(\Rightarrow c^{2}=a c+b\)…..(i)
Now, \(f(x)\) is differentiable at \(x=c\).
\(\Rightarrow(\) L.H.D. at \(x=c)=(\) R.H.D. at \(x=c)\)
\(\Rightarrow \mathop {\lim }\limits_{x \to {c^ – }} \frac{{f(x) – f(c)}}{{x – c}} = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{f(x) – f(c)}}{{x – c}}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to c} \frac{{{x^2} – {c^2}}}{{x – c}} = \mathop {\lim }\limits_{x \to c} \frac{{(ax + b) – {c^2}}}{{x – c}}\)
\( \Rightarrow \mathop {\lim }\limits_{x \to c} \frac{{{x^2} – {c^2}}}{{x – c}} = \mathop {\lim }\limits_{x \to c} \frac{{ax + b – (ac + b)}}{{x – c}}\) [Using (i)]
\( \Rightarrow \mathop {\lim }\limits_{x \to c} (x + c) = \mathop {\lim }\limits_{x \to c} \frac{{a(x – c)}}{{x – c}}\)
\(\Rightarrow \mathop {\lim }\limits_{x \to c} (x + c) = \mathop {\lim }\limits_{x \to c} a\)
\(\Rightarrow 2 c=a\) ……(ii)
From(i) and (ii), we have
\(c^{2}=2 c^{2}+b\)
\(\Rightarrow b=-c^{2}\)
Hence, \(a=2 c\) and \(b=-c^{2}\)

Summary

A function \(f(x)\) is said to be continuous at the point \(x=a\) if \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\). Thus, continuity of \(f(x)\) at \(x=a\) implies \(f(x)\) is defined at \(\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\). Similarly, \(f(x)\) is differentiable at \(x=c\), if \(\mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) – f\left( c \right)}}{{x – c}}\) exists finitely, if \(\mathop {\lim }\limits_{x \to {c^ – }} \frac{{f(x) – f(c)}}{{x – c}} = \mathop {\lim }\limits_{x \to {c^ + }} \frac{{f(x) – f(c)}}{{x – c}}\). The relationship between continuity and differentiability is that if a function is differentiable at a point, it is necessarily continuous. But, the converse may not be necessarily true. In simpler words, we can say that all differentiable functions are continuous, but not all continuous functions are differentiable.

Frequently Asked Questions (FAQs)

Q.1. What is the relation between the differentiability and continuity of a function?
Ans: The main relation between differentiability and continuity of a function can be stated as: All differentiable functions are also continuous functions. However, not all continuous functions can be claimed to be differentiable.

Q.2. Is continuity necessary for differentiability?
Ans: Any differentiable function, in particular, must be continuous throughout its domain. The opposite is not true: a continuous function does not have to be differentiable.
Example: The function \(f(x)=|x|\) is continuous at \(x=0\), but not differentiable at \(x=0\).

Q.3. What is the difference between continuous and differentiable?
Ans: The continuous function is a function for which the curve is single unbroken curve.
In contrast, if a function has a derivative, it is said to be differentiable.

Q.4. How do you calculate continuity and differentiability?
Ans: For checking the continuity of a function at \(x=a\) in its domain, check the right-hand limit, the left-hand limit, and the value of the function at point \(a\).
The function is said to be continuous if \(f(a)\) is defined and \(\mathop {\lim }\limits_{x \to {a^ – }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\)
Let \(f(x)\) be a real-valued function defined in an open interval \((a, b)\) and let \(c \in(a, b)\). Then, \(f(x)\) is said to be differentiable at \(x = c \Leftrightarrow \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) – f\left( c \right)}}{{x – c}}\) exists finitely.

Q.5. What does differentiable mean on a graph?
Ans: A function is differentiable at a point on the curve where a defined derivative exists. This indicates that the slope of the tangent line connecting the points on the left is approaching that of the tangent line connecting the points on the right.

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