Rolle's Theorem:

Statement: Let f be real valued function defined on closed interval [a, b] such that,

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

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

(iii) f(a) = f(b)

Then there is at least one value of C of x in (a, b) for which f'(c) = 0

Proof: Case I: f(x) is constant function in interval [a, b] then f'(x) = 0 for all x [a, b]

Hence, follows Rolle's theorem, we can say that f'(c) = 0 where a < c < b.

Case II: f(x) is not constant in interval [a, b] & sionce f(a) = f(b)


Let f(x) increases for x > a.
Since f(a) = f(b), so, function must increase to some value x = c & decreasing upto x = b.
At x = c function has maximum value.
Let h be small quantity, then,
    f(c + h) - f(c) < 0    &    f(c - h) - f(c) < 0


But if
Then Rolle's theorem cannot applicable b/c. f(x) is not differentiable at x = c
Only possible way for Rolle's theorem, when
    
f'(c) = 0    where a < c < b

Note: (i) Polynomial function is everywhere cont. & differentiable.

(ii) Exponential function, sine & cosine function are everywhere cont. & differentiable.

(iii) Logarithmic function is cont. & differentiable in its domain.

(iv) tanx is not cont. & differentiable at x = ± , ± , ...............

(v) |x| is not differentiable at x = 0.


Illustration: Verify Rolle's theorem for function f(x) = x³ - 3x² + 2x in interval [0, 2].

Ans: (a) f(x) is polynomial so, it is cont. & differentiable everywhere.

(b) f'(x) = 3x² - 6x + 2 clearly exista for all x (0,2)

(c) f(0) = 0,   f(2) = 2³ - 3(2)² + 2(2) = 0
f(0) = f(2)
So, All conditions of Rolle's theorem are satisfied.
So, there exist some c (0,2) such that f'(c) = 0
f'(c) = 3c² - 6c + 2 = 0    c = 1 ±


Lagrange's Mean Value Theorem:

Statement:

(i) f(x) is cont. in closed interval [a, b].

(ii) is differentiable in open interval (a, b)

Then there is at least one value c (a, b) such that
f'(c) =

Geometrical interpretation:


Let A, B be points on curve y = f(x) corresponding to x = a & x = b, so that
A = [a, f(a)]    &    B = [b, f(b)]
Slope of chord AB =
But slope of chord AB = f'(c), the slope of tangent to curve at x = c.
Accordin to LMVT, if a curve has tangent at each of its points then there is a point 'c' on this curve in between A & B, the tangent at which is paralled to chord AB.


Illustration: Find c of LMVT for which f(x) = in [1, 5].

Ans: (1) f(x) has definite & unique value of each x [1, 5]
So, every point in interval [1, 5] the value of f(x) is equal to limit of f(x).
f(x) is cont. in [1, 5]

(2) f'(x) = - exists for all x [1, 5]
f(x) is differentiable in (1, b)
So, there must br some c such that
f'(c) =
But f'(c) =


Application of derivative in determining the nature of roots of cubic polynomial:

Let f(x) = x³ + ax² + bx + c be given cubic polynomial.
f(x) = 0
f'(x) = 3x² + 2ax + b ....................................................... (i)
Let D = 4a² - 12b = 4(a² - 3b) be discriminate of eq. f'(x) = 0

Case I: If D < 0   f'(x) > 0 v x R
f(x) = - and


from fig. graph of y = f(x) cut x-axis only once. So, we have only real root (say x₀)
x₀ > 0 if c < 0 & x₀ < 0 if c > 0

Case II: If D > 0, f'(x) = 0
have two real roots.
f'(x) = 3(x - x₁)(x - x₂)
    f'(x) < 0, x (x₁, x₂)
    f'(x) > 0, x (- , x₁) U (x₂, )
f(x) would increase in (-, x₁) & (x₂, ) and would decrese in (x₁, x₂)
x = x₁ would be point of local maxima & x = x₂ would be point of local minima.


3 distict roots  x = , ,           One real root   x = & two imaginary root


One real root   x = 9 &         Three roots   x = , n₂, x₂
two imaginary roots.            (x₂ is repeated root)




Results:

(a) From I graph,   f(x₁) > 0, f(x₂) < 0
f(x₁) f(x₂) < 0, f(x) = 0 would have 3 real & distinct roots.

(b) Fromn II & III graph f(x₁) f(x₂) > 0, f(x) = 0 have one real & two imaginary roots.

(c) f(x₁ f(x₂) = 0, f(x) = 0 have 3 real roots but of root would be repeated.

Case III: D = 0, f'(x) = 3(x - x₁)⁰ where x₁ is root of f'(x).
f(x) = (x - x₁)³ + K
Then, f(x) = 0, has three real roots if K = 0
f(x) = 0 has one real root if K 0


Illustration: Find values of a so that x³ - 3x + a = 0 has three real & distinct roots.

Ans: Let f(x) = 3x³ - 3x + a
f'(x) = 3x² - 3 = 3(x - 1)(x + 1)
x₁ = 1, x₂ = - 1
f(x₁) = f(1) = a - 2
f(x₂) = f(- 1) = a + 2
Since roots would be real & distinct if
    f(1) f(- 1) < 0   (a - 2)(a + 2) < 0
- 2 < a < 2