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Mathematics. Session. Applications of Derivatives - 3. Session Objectives. Rolle’s Theorem Geometrical Meaning Lagrange’s Mean Value Theorem Geometrical Meaning Approximation of Differentials Class Exercise. Then, there is a point c in the open interval (a, b), such that.

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## Mathematics

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**Session**Applications of Derivatives - 3**Session Objectives**• Rolle’s Theorem • Geometrical Meaning • Lagrange’s Mean Value Theorem • Geometrical Meaning • Approximation of Differentials • Class Exercise**Then, there is a point c in the open interval(a, b), such**that Rolle’s Theorem • Let f(x) be a real function defined in the closed interval [a, b] such that • f(x) is continuous in the closed interval [a, b] • (ii) f(x) is differentiable in the open interval (a, b). • (iii) f(a) = f(b)**Geometrical Meaning**There will be at least one point with in [a, b] where tangent of the curve will be parallel to x-axis.**Example - 1**Verify Rolle’s theorem for the functionf(x) = x2 – 8x + 12 on the interval[2, 6]. Solution : We have f(x) = x2 – 8x + 12 (1) Given function f(x) is polynomial function. \ f(x) is continuous on [2, 6] (2) f'(x) = 2x – 8 exists in (2, 6) \ f(x) is differentiable in (2, 6)**\Three exists some such that f'(c) = 0**Solution (3) f(2) = 22 – 8 x 2 + 12 = 0 and f(6) = 62 – 8 x 6 + 12 = 0 \ f(2) = f(6) \ All the conditions of Rolle’s theorem is satisfied. Hence, Rolle’s theorem is verified.**Using Rolle’s theorem, find the points on the curve**where the tangent is parallel to x-axis. Function is differentiable in (0, 4). Example - 2 Solution: (1) Being a polynomial function, the given function is continuous on [0, 4].**All conditions of Rolle’s theorem are satisfied.**Con. Required point is (2, –4)**Then,there exists a real number**such that Lagrange’s Mean Value Theorem • Let f(x) be a function defined on [a, b] such that • it is continuous on [a, b]. • (ii) it is differentiable on (a, b).** Slope of the chord AB = Slope of the tangent at**Geometrical Meaning From the triangle AFB, By Lagrange’s Mean Value theorem,**(2) f'(x) = 2x – 3 exists in (-1, 2)**f(x) is differentiable in (-1, 2) Example - 3 Verify Lagranges Mean Value theorem forthe function f(x) = x2 – 3x + 2 on [-1, 2]. Solution : • The function f(x) being a polynomial function is continuous in [-1, 2].**So, there exists at least one such that**Solution Hence, Lagrange's mean value theorem is verified.**Using Lagrange’s mean value theorem, find the point on the**curve , where tangent is parallel to the chord joining (1, –2) and (2, 1). Function is differentiable in (1, 2). such that tangent is parallel to chord joining (1, –2) and (2, 1) Example - 4 Solution: (1) The function being a polynomial function is continuous on [1, 2].**As by the definition of**Approximation of Differentials Hence, for small increment in x, change in y will be**Using differentials, find the approximatevalue of**Example - 5 Solution :**Contd.**= 6 + 0.08 = 6.08**Using differentials, find the approximatevalue of**Example - 6 Solution :**So, there exists at least one where tangent is**parallel to chord joining (3, 0) and (5, 4). At x = 4 Required point is (4, 1) Solution**Using differentials, find the approximate value ofup to 3**places of decimals. Class Exercise - 5 Solution :

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