If a function is continuous in [a, b] then it attains all the values between f (a) and f (b) including f (a) and f (b)

__Rolle’s Theorem:__

It is one of the most fundamental theorem of Differential calculus and has far reaching consequences. It states that if y = f (x) be a given function and satisfies,

∎ f(x) is continuous in [a , b]

∎ f(x) is differentiable in (a , b )

∎ f(a) = f(b)

Then f'(x) = 0 at least once for some x∈(a , b)

If f(x) satisfies the conditions of Rolle’s theorem in [a, b] ; it’s derivative would vanish at least once in (a , b)

$ \displaystyle A \equiv (a , f(a)) \, , \, B \equiv (b , f(b)) $

as f(a) = f(b) (third condition of Rolle’s theorem)

Slope of line AB = 0

We would have at least one point belonging to (a, b) so that tangent drawn to the curve at that point would be parallel to the line AB

__Applications of Rolle’s Theorem:__

∎ If y = f (x) satisfies the Rolle’s theorem in [a , b], then f'(x) = 0 for some x∈ (a, b). As any solution of f'(x)=0 would give us a root of f'(x) = 0 , hence we can say that at least one root of f'(x)= 0 would belong to (a, b) if f(x) satisfies all conditions of Rolle’s Theorem.

∎ Let x = a and x = b be the roots of f (x) = 0 and y = f (x) satisfies the condition of Rolle’s theorem in [a, b]. Here f (a) = f (b) = 0. Hence we can say that between two roots of f (x) = 0 at least one root of f'(x)=0 would lie.

∎ If y = f(x) be a polynomial function of degree n. If f (x) = 0 has real roots only, then, f'(x) = 0 , f”(x) = 0 ,…f^{n-1}(x)= 0 would have real roots. It is in fact the general version of application no.2, because if f (x) = 0 have all real roots, then between two consecutive roots of f(x) = 0, exactly one root of f'(x)= 0 would lie.

### Lagrange’s Mean Value Theorem

This theorem is in fact the general version of Rolle’s theorem.

It says that if y = f(x) be a given function which is;

∎ Continuous in [a , b]

∎ Differentiable in (a , b)

Then,

$\displaystyle f'(x) = \frac{f(b)-f(a)}{b-a} $ atleast once for some x∈(a, b)

Let A ≡ (a , f (a)) and B ≡ (b , f (a))

Slope of Chord $ \displaystyle AB = \frac{f(b)-f(a)}{b-a} $

As f'(x) gives us the slope of tangent at the point (x , y), this theorem simply says that there will be at least one point ∈ (a, b) e.g. (points c_{1} , c_{2} and c_{3}) such that tangent drawn to the curve at this point would be parallel to the chord connecting points A and B.

We can have one more interpretation, i.e. is the instantaneous rate of change of f(x) and $ \displaystyle \frac{f(b)-f(a)}{b-a} $ gives us the average rate of change of f (x) over [a , b] .

This theorem simply says that average rate of change of the function over a given interval would be equal to instantaneous rate of change of function on at least one point of that interval.

It is a well known fact in physics you must have learnt that average velocity of a particle over an interval is equal to instantaneous velocity of the particle at some point of that interval.

Illustration : If 2a + 3b + 6c = 0 then prove that the equation ax^{2} + bx + c = 0 would have at least one root in (0, 1); a , b , c ∈ R

Solution: Let $ \displaystyle f'(x) = ax^2 + bx +c $

$ \displaystyle f(x) = \frac{a x^3}{3} + \frac{bx^2}{2} + cx + d $

f (0) = d

Also, $ \displaystyle f(1) = \frac{2a + 3b + 6c}{6} = d $

Hence all the conditions of Rolle’s theorem are satisfied in [0, 1]

So, f’ (x) = 0 for atleast one value in (0, 1)

Illustration : If p (x) = 51x^{101} – 2323x^{100} – 45x + 1035 , using Rolle’s Theorem, prove that atleast one root lies between (45^{1/100} , 46)

Solution:

Let $ \displaystyle g(x)=\int p(x) dx $

$ \displaystyle = \frac{51 x^{102}}{102}- \frac{2323 x^{101}}{101}-\frac{45 x^2}{2}+1035 x + c $

$ \displaystyle = \frac{x^{102}}{2}- 23 x^{101}-\frac{45 x^2}{2}+1035 x + c $

Now g (45^{1/100}) $ \displaystyle = \frac{1}{2}(45)^{102/100}- 23 (45)^{101/100}-\frac{45}{2} (45)^{2/100}+1035 (45)^{1/100} + c = c $

g (46) $ \displaystyle = \frac{(46)^{102}}{2}- 23 (46)^{101}-\frac{45 (46)^2}{2}+1035 (46) + c = c $

So g'(x) = p (x) will have atleast one root in given interval

Illustration : Prove that if 2a_{0}^{2} < 15a , then all roots of x^{5} – a_{0} x^{4} + 3ax^{3} + bx^{2} + cx + d = 0 can ‘ t be real. It is given that a_{0}, a, b, c, d ∈ R.

Solution: Let f (x) = x^{5} – a_{0} x^{4} + 3ax^{3} + bx^{2} + cx + d

f’ (x) = 5x^{4} – 4a_{0} x^{3} + 9ax^{2} + 2bx + c

f” (x) = 20x^{3} – 12a_{0} x^{2} + 18ax + 2b

f”’ (x) = 60x^{2} – 24a_{0} x + 18a

= 6(10x^{2} – 4a_{0}x + 3a)

Now, discriminant = 16a_{0}^{2} – 4. 10. 3a = 8 (2a_{0}^{2} -15a) < 0

as 2a_{0}^{2} – 15a < 0 is given .

Hence the roots of f”'(x) = 0 can not be real.

And therefore all the

roots of f(x) = 0 will not be real.

Exercise :

(i) Show that there lies a point on the curve f(x) = x^{2} – 4x +3 between (1 , 0) and (3, 0) where tangent drawn is parallel to x-axis. Find its coordinates.

(ii) Show that there lies a point on the curve f(x) = x(x +3)e^{-x/2} in the interval (-3, 0) where tangent drawn is parallel to x-axis.

(iii) By considering the function f(x) = (x – 2) logx, show that the equation x logx = 2 – x is satisfies by at least one value of x lying between 1 and 2.