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| Rolle's Theorem and Mean Value Theorem |
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| Let f be a real valued function in [a,b] such that |
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 f is continuous in [a,b]. |
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 f is differentiable in (a,b). |
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| Geometrical meaning |
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| Let A (a,f (a)) and B (b,f (b)) be two points on the graph of f (x) such
that f(a) = f(b), then $ c
Î (a, b) such that the tangent at P(c, f(c)) is
parallel to x - axis. |
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| Note 1: We cannot obtain c if any one of the conditions of Rolle's theorem are not satisfied. |
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| Note 2: The value of c need not be unique. |
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| Example: |
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| Verify Rolle's theorem for the function |
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| f (x) = x2 - 8x + 12 on (2, 6) |
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| Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied. |
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| f (2) = 22 - 8 (2) + 12 = 0 |
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| f (6) = 36 - 48 + 12 = 0 |
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| Therefore (iii) condition is satisfied. |
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 Rolle's theorem is applicable for the given function f (x). |
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\ There must exist c  (2, 6) such that f '(c) = 0 |
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| f '(x) = 2x - 8 |
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 Rolle's theorem is verified. |
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| Let f (x) be a function defined on [a, b]. |
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| Step 1: |
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| Show that the function is continuous in the given interval. Some known standard functions which are continuous, can be mentioned directly. |
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| Step 2: |
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| Differentiate f (x) and examine if f '(x) is defined at every point in the open interval (a, b). |
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| Step 3: |
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| Check if f (a) = f (b) |
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| If all the above condition are satisfied, then Rolle's theorem is applicable else the Rolle's theorem is not applicable. |
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| If Rolle's theorem is applicable, solve f '(c) = 0. Show that one of these roots lie in the open interval (a, b). |
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