Unit-3 Lecture 1 PDF

## Rolli's Theorem **If a function f is continuous on [a,b], differentiable on (a,b) and f(a) = f(b); then there exists a number c ∈ (a,b) such that f’(c) = 0** **Geometrical Interpretation:** * f’(c) = 0 at x = c, tangent is parallel to x-axis. * Thus can exist more than one such points, when th...

## Rolli's Theorem **If a function f is continuous on [a,b], differentiable on (a,b) and f(a) = f(b); then there exists a number c ∈ (a,b) such that f’(c) = 0** **Geometrical Interpretation:** * f’(c) = 0 at x = c, tangent is parallel to x-axis. * Thus can exist more than one such points, when the tangent is parallel to x-axis. **Example:** * Find a value of c satisfying the conclusion of Rolli’s Theorem for f(x) = 1/3 x^3 - 3x^2 + 2x + 2 on [0,1]. * Sol: f(x) being a polynomial is continuous & differentiable everywhere on [0,1] as well * Now f(0) = 2 and f(1) = 2 * By Roll’s Theorem; f’(x) = 0 for c ∈ (0,1) * 3c^2 - 6c + 2 = 0 * c = (1 ± √3)/3 * But c = (1 + √3)/3 ∉ (0,1) * c = (1 - √3)/3 = 0.42265 **Example:** * Find a value of c satisfying the conclusion of Rolli’s Theorem for f(x) = sin x in [0, π] * Sol: f(x) being a trigonometric function is continuous in [0, π] also differentiable in (0, π) * f(0) = 0 = f(π) * By Roll’s Theorem, there exists c ∈ (0, π) such that f’(c) = 0 * Cos c = 0 * c = π/2 **Example:** * Find a value of c satisfying the conclusion of Rolli’s Theorem for f(x) = 1/3 x^3 - x^2 + 1 in [0, 1]. * Sol: f(x) being a polynomial is continuous on [0, 1] and differentiable on (0, 1). * Also f(0) = 1 * By Roll’s Theorem, there exists c ∈ (0, 1) such that f’(c) = 0 * c^2 - 2c + 1 = 0 * c = 1 ± 1/√3 * But c = 1 + 1/√3 ∉ (0, 1) * c = 1/√3 ## Mean Value Theorem **If the function f is continuous on [a,b] and differentiable on (a,b) then there exists a number c ∈ (a,b) such that f’(c) = f(b) - f(a) / b - a** **Geometrical Interpretation:** * There exist at least one point c between A & B. The Tangent is parallel to the chord AB. **Example:** * Find the value of c satisfying the conclusion of Mean Value Theorem for f(x) = x^3 - x^2 + x + 1 on [0,2] * Sol: f(x) being a polynomial is continuous on [0,2] and differentiable on (0, 2) * By MVT, there exists c ∈ (0, 2) for which f’(c) = f(2) - f(0) / 2 - 0 = 3 - 1 / 2 = 1 * 3c^2 - 2c - 1 = 1 * 3c^2 - 2c - 2 = 0 * c = (2 ± √(4 + 24)) / 3 = 1 ± √7 /3 * But c = (1 - √7)/3 ∉ (0,2) * c = (1 + √7)/3 **Example:** * Find the value of c satisfying the conclusion of MVT for f(x) = 2x^2 + 3x + 4 in [1,2], Ans = 3/2 * Find the value of c satisfying the conclusion of MVT for f(x) = x(x-1)(x- 2) in [0, 1/2], ans = 1/3 * Sol: f(x) being a polynomial function is continuous on [0, 1/2] and differentiable on (0, 1/2) * Thus, there exists c ∈ (0, 1/2) such that f’(c) = (f(1/2)- f(0))/ 1/2 - 0 = (-3/8 - 0) / 1/2 = -3/4 * 3c^2 - 6c + 2 = -3/4 * 12c^2 - 24c + 8 = -3 * 12^2 - 24c + 5 = 0 * c = 6 ± √21 / 6 * But c = 6 + √21 / 6 ∉ (0, 1/2) * c = 6 - √21 / 6