Intermediate Value Theorem

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 ,…fn-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 c1 , c2 and c3) 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 ax2 + 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) = 51x101 – 2323x100 – 45x + 1035 , using Rolle’s Theorem, prove that atleast one root lies between (451/100 , 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 (451/100) $ \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 2a02 < 15a , then all roots of x5 – a0 x4 + 3ax3 + bx2 + cx + d = 0 can ‘ t be real. It is given that a0, a, b, c, d ∈ R.

Solution: Let f (x) = x5 – a0 x4 + 3ax3 + bx2 + cx + d

f’ (x) = 5x4 – 4a0 x3 + 9ax2 + 2bx + c

f” (x) = 20x3 – 12a0 x2 + 18ax + 2b

f”’ (x) = 60x2 – 24a0 x + 18a

= 6(10x2 – 4a0x + 3a)

Now, discriminant = 16a02 – 4. 10. 3a = 8 (2a02 -15a) < 0

as 2a02 – 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) = x2 – 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.

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