# 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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