# application-of-derivatives-3

__:__

**Rolle's Theorem**__: Let f be real valued function defined on closed interval [a, b] such that,__

**Statement**(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

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

**Note**(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.

__: Verify Rolle's theorem for function f(x) = x__

**Illustration**^{3}- 3x

^{2}+ 2x in interval [0, 2].

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

(b) f'(x) = 3x

^{2}- 6x + 2 clearly exista for all x (0,2)

(c) f(0) = 0, f(2) = 2

^{3}- 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) = 3c

^{2}- 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.

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

**Illustration**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) = x

^{3}+ ax

^{2}+ bx + c be given cubic polynomial.

f(x) = 0

f'(x) = 3x

^{2}+ 2ax + b ....................................................... (i)

Let D = 4a

^{2}- 12b = 4(a

^{2}- 3b) be discriminate of eq. f'(x) = 0

__Case I__: If D < 0 f'(x) > 0

f(x) = - and

from fig. graph of y = f(x) cut x-axis only once. So, we have only real root (say x

_{0})

x

_{0}> 0 if c < 0 & x

_{0}< 0 if c > 0

__Case II__: If D > 0, f'(x) = 0

have two real roots.

f'(x) = 3(x - x

_{1})(x - x

_{2})

f'(x) < 0, x (x

_{1}, x

_{2})

f'(x) > 0, x (- , x

_{1}) U (x

_{2}, )

f(x) would increase in (-, x

_{1}) & (x

_{2}, ) and would decrese in (x

_{1}, x

_{2})

x = x

_{1}would be point of local maxima & x = x

_{2}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 = , n

_{2}, x

_{2}

two imaginary roots. (x

_{2}is repeated root)

Threen real roots x = n

_{1}, x_{2},(x

_{1}is repeated roots)__:__

**Results**(a) From I graph, f(x

_{1}) > 0, f(x

_{2}) < 0

f(x

_{1}) f(x

_{2}) < 0, f(x) = 0 would have 3 real & distinct roots.

(b) Fromn II & III graph f(x

_{1}) f(x

_{2}) > 0, f(x) = 0 have one real & two imaginary roots.

(c) f(x

_{1}f(x

_{2}) = 0, f(x) = 0 have 3 real roots but of root would be repeated.

__Case III__: D = 0, f'(x) = 3(x - x

_{1})

^{0}where x

_{1}is root of f'(x).

f(x) = (x - x

_{1})

^{3}+ K

Then, f(x) = 0, has three real roots if K = 0

f(x) = 0 has one real root if K 0

__: Find values of a so that x__

**Illustration**^{3}- 3x + a = 0 has three real & distinct roots.

Ans: Let f(x) = 3x

^{3}- 3x + a

f'(x) = 3x

^{2}- 3 = 3(x - 1)(x + 1)

x

_{1}= 1, x

_{2}= - 1

f(x

_{1}) = f(1) = a - 2

f(x

_{2}) = f(- 1) = a + 2

Since roots would be real & distinct if

f(1) f(- 1) < 0 (a - 2)(a + 2) < 0

- 2 < a < 2