Rolle’s Theorem: The mean value theorem has the utmost importance in differential and integral calculus. Rolle’s theorem is a special case of the mean value theorem. While in the mean value theorem, the minimum possibility of points giving the same slope equal to the secant of endpoints is discussed, we explore the tangents of slope zero of functions in Rolle’s theorem.

Let us familiarise ourselves and learn more about Rolle’s theorem in this article.

Rolle’s Theorem Statement

If a function \(f(x)\) defined on closed interval \([a, b]\) is:

1. continuous on closed interval \([a, b]\)
2. derivable on open interval \((a, b)\)
3. \(f(a)=f(b)\)

then there exists at least one real number \(c,\) between \(a\) and \(b, (a<c<b),\) such that \(f'(c)=0.\)

Graphical Visualisation of Rolle’s Statement

One can easily observe that the given curve satisfies all Rolle’s conditions and possesses the point \(C(c,f(c)),\) such that: 

1. the function’s derivative is zero (as tangent drawn at \(c\) is parallel to the \(x-\)axis) at \(x=c.\)
2. there is exactly one root for the derivative equal to zero between \(x=a\) and \(x=b.\)

In some cases, we can have multiple roots of derivative equal to zero between \(x=a\) and \(x=b.\)

The curve above satisfies all Rolle’s conditions again. Hence, the theorem can be applied between \(a\) and \(b.\) Here, multiple roots such as \({c_1},{c_2},{c_3}\)  can be observed where \(f'(x)=0.\)

The theorem only comments on the least possible number of roots of \(f'(x)=0,\) at least one.

Geometrical Interpretation

The theorem states that between two points with equal ordinates on the \(f(x)\) graph, there is at least one point where the tangent is parallel to the \(x\)-axis. In previous examples, one can see that if we draw a tangent at turning points of the curve, its slope is zero, and the tangent is parallel to the \(x-\)axis.

Algebraic Interpretation 

Between two zeros \(\alpha \) and \(\beta \) of \(f(x),\) there exists at least one zero of \(f'(x).\) Let’s try to understand this using some graphs.

In this graph, we can observe that \(a\) and \(b\) are the roots of the function \(f(x),\) and between them lies \(c.\) This is the root of the derivative \(f’x=0\) at \(x=c.\)

Though this curve has only one root for \(f'(x)=0,\) there can be more than one root for functions, as shown in the graph shown below.

In this graph of the function \(f(x),\) between two roots of \(f(x)\) at \(a\) and \(b,\) there exist several roots of \(f'(x)=0,\) at five different points on the curve. Hence, we can conclude that the minimum number of roots for derivatives is based on information about the roots of the function \(f(x)=0.\)

Let’s have a look at the following graph:

Here, \(f(x)\) has four distinct zeros as \(a, b, c,\) and \(d,\) and the number of zeros of derivatives is equal to \(7.\)

We can conclude that if a continuous and differentiable function \(f(x)=0\) has \(‘n’\) real roots, then its derivative \(f'(x)=0\) will have at least \(n-1\) real roots. But we cannot comment about zeroes of \(f(x)\) if the number of zeros of \(f'(x)\) is known. Let us look at some more examples.

In the first graph, one should observe that there are three zeroes for the \(f'(x)\) but no zeroes at all for the \(f(x)\) and in the second graph \(f'(x)\) has three zeros, and \(f(x)\) has two. Hence, the number of roots for \(f(x)\) cannot be calculated when the roots of \(f'(x)\)  are known.

Rolle’s Theorem Proof

Since \(f(x)\) is continuous and differentiable on closed intervals \([a, b], f(x)\) is bound in its domain, attaining its maximum \((M)\) and minimum \((m)\) value. 

Case 1: Maximum and minimum value are equal \((M=m)\)

In this case \(f(x)\) is a constant, hence, \(f'(x)=0,\) for all value of \(x.\)

Case 2: Maximum \((M)\) and minimum \((m)\) are not equal \((M≠m)\) in \((a,b)\)

If \(m\) and \(M\) are distinct, then there must be one point \(c,\) where \(f(x)\) attains its maximum value.

Now at point \(C,f(c + h) – f(c) \le 0\) and \(f(c – h) – f(c) \le 0\)

\(\therefore \frac{{f(c + h) – f(c)}}{h} \le 0\) and \(\frac{{f(c – h) – f(c)}}{{ – h}} \ge 0\)

\(\mathop {\lim }\limits_{h \to 0} \frac{{f(c + h) – f(c)}}{h} \le 0\) and \(\mathop {\lim }\limits_{h \to 0} \frac{{f(c – h) – f(c)}}{{ – h}} \ge 0\)

So, the right-hand derivative and the left-hand derivative at \(C\) have different signs (making it non-differentiable at \(C\)), which is a contradiction because we know that \(f(x)\) is differentiable.

Hence, \(f'(c)\) must be equal to zero.

Solved Problems – Rolle’s Theorem:

Q.1. Verify Rolle’s theorem for \(f(x) = {x^2} – 3x + 2,x \in [1,2]\)
Ans: 1. \(f(x)\) is a quadratic polynomial which is always continuous and differentiable for all real values.
2. Further \(f(1) = 1 – 3 + 2 = 0\) and \(f(2) = 4 – 6 + 2 = 0,\)
So, all the Rolle’s conditions are satisfied and as per Rolle’s, there should be at least one \(‘c’\) for which
the derivative of \(f(x)\) should be zero in \([1, 2],\) Now
\(f(x) = {x^2} – 3x + 2\)
\({f^\prime }(x) = 2x – 3\)
\({f^\prime }(c) = 2c – 3 = 0 \Rightarrow c = \frac{3}{2}\)
We can observe that \(c\) is lying between \(1\) and \(2,\) which validates Rolle’s statement.

Q.2. If \(2a+3b+6c = 0\) then prove that the equation
\(a{x^2} + bx + c = 0\)has at least one root in \((0, 1).\)
Ans: Assume that there exists a function \(f(x)\) such that \({f^\prime }(x) = a{x^2} + bx + c\)
So, we can find \(f(x)\) by integrating \(f’(x):\)
\(\int {{f^\prime }} (x)dx = \int {\left( {a{x^2} + bx + c} \right)} dx\)
\( \Rightarrow f(x) = \frac{{a{x^3}}}{3} + \frac{{b{x^2}}}{2} + cx + d\) where d is integration constant. Now,
\(f(1) = \frac{{2a + 3b + 6c}}{6} + d\)
\( \Rightarrow f(1) = d\)
Also \(f(0) = d\)
As we can see, \(f(x)\) is a cubic polynomial, which is always continuous and differentiable, and \(f(1)\) is equal to \(f(0).\)
It satisfies all the conditions for the application of ROLLE’S theorem in \([0, 1].\)
So, applying Rolle’s on \(f(x),\) we can conclude that there is at least one \(c\) for which f’(c) = \(0\) for \(x\) belonging to the interval \([0, 1].\)
Hence, we can say that,
\(a{x^2} + bx + c = 0\) has at least one root \(x\) belonging to the interval \([0, 1].\)

Q.3. Let \(f(x)\) be a differentiable function and \(g(x)\) be twice differentiable. Zeros of \(f(x), g’(x)\) be \(a\) and \(b\) respectively \((a<b).\) Show that there exists at least one root of the equation \(f’(x)g’(x)+f(x)g”(x)=0\) on the interval \((a, b).\)
Ans: Assume that \({H^\prime }(x) = {f^\prime }(x){g^\prime }(x) + f(x){g^{\prime \prime }}(x)\) we can find \(H(x)\) from given relation:
\({H^\prime }(x) = {f^\prime }(x){g^\prime }(x) + f(x){g^{\prime \prime }}(x)\)
\(\frac{{dH(x)}}{{dx}} = \frac{{df(x){g^\prime }(x)}}{{dx}}\)(x)dx
\( \Rightarrow d(H(x) = d((f(x)g(x))\)
\(\Rightarrow\int d \left( {H(x) = \int d (f(x)g(x))} \right.\)
\( \Rightarrow H(x) = f(x)g(x) + c\)
Given that \(f(a)=0\) and \(g’(b)=0,\)
So , \(H(a)=c\) and \(H(b)=c\)
As it is given that \(f(x)\) is a differentiable function and \(g(x)\) is twice differentiable,
so \(H(x)\) will also be continuous and differentiable.
Further, \(H(a)=H(b)\)
\(H(x)\) satisfies all the conditions for the application of Rolle’s on \((a, b).\)
Hence, \(H’(x)=f’(x)g’(x)+f(x)g”(x)\) will have at least one root in \((a, b).\)

Q.4. If Rolle’s theorem is applicable to the function \(f(x) = \ln (x)/x,(x > 0)\) over the interval \([a, b]\) where \(a\) and \(b\) are integers, then value of \({a^2} + {b^2}\) can be
(A) \(20\) (B) \(25\) (C) \(45\) (D) \(10\)
Ans: Given that Rolle’s is applicable over \([a, b]\)
Hence,
\(f(a) = f(b)\)
\(\frac{{\ln lna}}{a} = \frac{{\ln lnb}}{b}\)
\(bln(a) = a\ln ln(b)\)
\(\ln ln\left( {{a^b}} \right) = ln\left( {{b^a}} \right)\)
\({a^b} = {b^a}\)
Also,
\(f(x) = \frac{{\ln (x)}}{x}\)
\({f^\prime }(x) = \frac{{1 – ln(x)}}{{{x^2}}}\)
\(f'(x)\) is zero at \(x=e.\) So a may be either \(1\) or \(2,\) and \(b\) may be equal to \(3\) onwards. By hit and trial, we can observe that if \(a=2\) and \(b=4\) it satisfies the condition \(a\) raise to the power \(b\) equals to braise to the power a equal to \(16.\)
For \(a=2,\)
\(f(a) = \frac{{\ln (2)}}{2}\)
And for \(b=4\)
\(f(b) = \frac{{\ln ln(4)}}{4}\) which simplifies to \(\frac{{ln(2)}}{2}\) giving us \(f(a) = f(b).\)
Hence, the sum of squares of \(a\) and \(b\) can be equal to \(20.\)
Furthermore:
\(25 = {1^2} + {(\sqrt {24} )^2}\) or \({2^2} + {(\sqrt {21} )^2}\) (Both rejected as \(b\) is not an integer)
\(45 = {1^2} + {(\sqrt {44} )^2}\) or \({2^2} + {(\sqrt {41} )^2}\) (Both rejected as \(b\) is not an integer)
\(10 = {1^2} + {(3)^2}\) where \(a\) and \(b\) both are integers but
\(\frac{{ln(1)}}{1} \ne \frac{{ln(3)}}{3}\) so, this also gets rejected.
Hence, the only choice is option (A).

Q.5. If \(f(x)\) and \(g(x)\) are differentiable functions for \(0≤x≤23\) such that \(f(0)=2, g(0)=0, f(23)=22, g(23)=10,\) then show that \(f’(x)=2g’(x)\) for at least one \(x\) in the interval \((0,23).\)
Ans: Let’s assume a function \(H(x)\) such that
\(H(x)=f(x)-2g(x)\)
So, \(H(0)= f(0)-2g(0)\)
\(=2-0\)
\(=2\)
And,
\(H(23)= f(23)-2g(23)\)
\(=22-20\)
\(=2\)
Hence, \(H(0)=H(23)\) and \(H(x)\) is continuous and differentiable over \([0, 23]\) as it is the sum of continuous and differentiable functions \(f(x)\) and \(g(x).\)
Applying ROLLE’s on \(H(x)\) in the interval \([0, 23],\)
\(H’(x)=0\) must have at least one root in \((0, 23).\)
Hence, \(f’(x)-2g’(x)=0\) is true for at least one \(x\) over the interval \((0, 23).\)

Summary

Rolle’s theorem has been proved as an important tool in finding possibilities of roots of derivatives. In general, for a continuous and derivable function with known roots, Rolle’s theorem is preferred for commenting about the existence of stationary points. Having learnt to prove Rolle’s theorem through examples, we have learnt to solve problems by its application.

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Frequently Asked Questions on Rolle’s Theorem

Q.1. Is Rolle’s theorem MVT?
Ans: Rolle’s theorem is a particular case of MVT. In the case of the mean value theorem, the interval in which it is applied does not need to have the same functional value at endpoints. Whereas in case of Rolle’s, functional value at endpoints for the interval \([a, b]\) is considered equal, i.e., \(f(a)=f(b).\)

Q.2. What are the three conditions of Rolle’s theorem?
Ans: To apply ROLLE’S on a function \(f(x),\) it should satisfy the following three conditions:
1. \(f(x)\) should be continuous on the closed interval \(a, b\)
2. \(f(x)\) should be derivable on the open interval \((a, b)\) and,
3. \(f(a)=f(b)\)

Q.3. Is the converse of Rolle’s theorem true?
Ans: Converse of Rolle’s theorem is not always true. We can have several counterexamples where the converse will not hold. For instance, consider the following curve:

In this curve, one can observe that there are two roots of \(F(x)=0\) in \(x\) belonging to \([a, b].\) Also, \(f(a)\) and \(f(b)\) are equal, yet, the function is not continuous and differentiable. Hence, the converse doesn’t hold.

Q.4. What is the difference between MVT and Rolle’s theorem?
Ans: As per statement of MVT, for a function \(f(x)\) which is continuous in closed interval \(\left[ {a,\,b} \right]\) and differentiable in open interval \((a, b),\) then there exists a value \(‘c’\) lying on the interval \(\left( {a,\,b} \right)\) such that
\(f’(c)=(f(b)-f(a))/ (b-a);\)
Now if \(f(a)\) and \(f(b)\) are equal then
\(f’(c)=0.\) This is what Rolle’s explains. There will exist at least one \(‘c’\) for which the derivative is equal to zero. The reason why Rolle’s is considered a particular case of MVT.

Q.5. Who is Rolle’s theorem named for?
Ans: Michel Rolle, a French mathematician, is best known for Rolle’s theorem \((1691).\)