Rolle's Theorem

Rolle's Theorem in Calculus

1. Welcome you all ! Dr. Shilpa M Assistant Professor of Mathematics MTI, Thrissur - 20

2. Rolle’s Theorem Rolle's theorem  states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(x) = 0 for some x with a ≤ x ≤ b.

3. Geometrical representation

4. In  calculus ,  Rolle's theorem  or  Rolle's lemma  essentially states that any real-valued  differentiable function  that attains equal values at two distinct points must have at least one  stationary point  somewhere between them—that is, a point where the first derivative (the slope of the tangent line to the graph of the function) is zero. The theorem is named after  Michel Rolle .

5. Example   Show that   f(x )=4x5+x3+7x−2f(x)=4x5+x3+7x−2 has exactly one real root . Solution : we know that since f(x)f(x) is a 5 th  degree polynomial it will have five roots . We have to prove is that only one of those 5 is a real number and the other 4 must be complex roots.

6. We now need to show that this is in fact the only real root. To do this we’ll use an argument that is called contradiction proof. What we’ll do is assume that f(x)f(x) has at least two real roots. This means that we can find real numbers  aa  and bb (there might be more, but all we need for this particular argument is two) such that f(a)=f(b)=0f(a)=f(b)=0. But if we do this then we know from Rolle’s Theorem that there must then be another number cc such that f′(c)=0f′(c)=0. This is a problem however. The derivative of this function is, f′(x)=20x4+3x2+7

7. Because the exponents on the first two terms are even we know that the first two terms will always be greater than or equal to zero and we are then going to add a positive number onto that and so we can see that the smallest the derivative will ever be is 7 and this contradicts the statement above that says we MUST have a number  c  such that f′(c)= 0. We reached these contradictory statements by assuming that  f(x)  has at least two roots. Since this assumption leads to a contradiction the assumption must be false and so we can only have a single real root.

8. Counter examples 1. Consider  f(x)={x} ({x} is the fractional part function ) on the closed interval [0,1 ]. In this case, the Rolle’s theorem fails because the function f(x) has a discontinuity at x=1 (that is, it is not continuous everywhere on the closed interval  [0,1].)

9. 2. The linear function f(x)=x is continuous on the closed interval [0,1] and differentiable on the open interval (0,1). The derivative of the function is everywhere equal to 1 on the interval. So the Rolle’s theorem fails here. This is explained by the fact that the 3rd condition is not satisfied (since f(0)≠f(1).)

10. THANK YOU !!!