Rolles theorem, Mathematics

Assignment Help:

Rolle's Theorem

 Assume f(x) is a function which satisfies all of the following.

1. f(x) is continuous in the closed interval [a,b].

2. f(x) is differentiable in the open interval (a,b).

3. f(a)  = f(b)

So, there is a number c as a < c < b and f′(c) = 0. Or, though f(x) has a critical point in (a,b).

 


Related Discussions:- Rolles theorem

Calculate the height of the tunnel and the perimeter, The adjoining figure...

The adjoining figure shows the cross-section of a railway tunnel. The radius of the tunnel is 3.5m (i.e., OA=3.5m) and ∠AOB=90 o . Calculate : i.       the height of the

Cone - three dimensional spaces, Cone - Three dimensional spaces The be...

Cone - Three dimensional spaces The below equation is the general equation of a cone. X 2 / a 2 + y 2 /b 2 = z 2 /c 2 Here is a diagram of a typical cone. Not

AREA, How do you find the distributive property any faster?

How do you find the distributive property any faster?

VECTOR, the sum of the vector QR, -SR, TQ and 2ST is?

the sum of the vector QR, -SR, TQ and 2ST is?

Describe adding and subtracting square roots, Describe Adding and Subtracti...

Describe Adding and Subtracting Square Roots? To add or subtract square roots, the radicands must be the same. If the radicands are the same, add/subtract the coefficients (the

Solid mensuration, given dimensions: 130cm, 180cm, and 190cm is to be divid...

given dimensions: 130cm, 180cm, and 190cm is to be divided by a line bisecting the longest side shown from its opposite vertex. what''s the area adjacent to 180cm? ;

Course 2 chapter 1 ratios and propotional reasoning, find the unit rate. Ro...

find the unit rate. Round to the nearest hundredth in necessary 325 meters in 28 seconds

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd