Rolles Theorem

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

Rolles theorem: Let 'f' be a real valued function defined on the closed interval[a,b] such that
1) it is continuous on the closed interval [a,b],
2) it is differentiable on the open interval (a,b),
3) f(a) = f(b).
Then there exists a real number c $\in$ (a,b) such that f '(c) = 0.

Geometric representation of Rolle's theorem
Let f(x) be the real valued function defined on [a,b] such that the curve y =f(x) is a continuous curve between the points A(a, f(a)) and B(b,f(b)) and it is possible to draw a unique tangent at every point on the curve y between A and B. Also, the coordinates of the end points of the interval [a,b] are equal. then, there exists at least one point (c,f(c)) lying between A and B on the curve y = f(x) where tangent is parallel to the X-axis.
Rolles theorem
Algebraic interpretation Of Rolle's theorem
Let f(x) be a polynomial function with a and b as its roots. As f(x) is a polynomial function so it is continuous and differentiable. And a and b are the roots of the function so f(a) = f(b) = 0. So f(x) satisfies all the conditions of Rolle's theorem. So there exists c such that c $\in$ (a,b) such that f '(c) = 0 or f'(x) = for x = c.
Between any two roots there is always a root of its derivative f'(x).

Examples on Rolles Theorem

Check the Rolle's theorem
1) f(x) = tan(x) on [0,$\pi$]
Solution : f(x) = tan(x)
As we know that the tangent function is not continuous in $\pi$/2 and $\pi$/2 lies in the interval of [0,$\pi$].
So for this function Rolle's theorem is not applicable here.

Verify the Rolle's theorem
1) f(x) = $x^{2} - 5x + 6 $ on the interval [2,3]
Solution : f(x) = $x^{2} - 5x + 6 $
As the given function is polynomial so it is continuous and differentiable in the interval given.
f(2) = $2^{2} - 5(2) + 6 $
= 4 - 10 + 6
f(2) = 0
f(3) = $3^{2} - 5(3) + 6 $
= 9 - 15 + 6
f(3) = 0
So f(2) = f(3)
As all the conditions of the Rolle's theorem are satisfied so there exists some 'c'.
f(x) = $x^{2} - 5x + 6 $
f'(x) = 2x -5
f'(c) = 2c - 5
0 = 2c - 5
2c = 5
c = 5/2 = 2.5
Hence, Rolle's theorem is verified.

12th grade math


Russia-Ukraine crisis update - 3rd Mar 2022

The UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.