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| Langrange's Mean Value Theorem |
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| Let f be 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 f (x), then |
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| Note: The value of c obtained need not be unique. |
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| Step 1: |
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| Show the function f (x) is continuous on the closed interval [a, b]. |
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| Step 2: |
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Find f '(x) and examine if it is defined at every point on the open interval (a, b). If f '(x) is defined for all x  (a, b), then the function is differentiable. |
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| Step 3: |
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| If the above condition are satisfied, then Mean Value Theorem is applicable. |
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| Step 4: |
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| If Mean value theorem is applicable, solve the equation |
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| Show that one of the roots lie in the open interval (a, b). This verifies the Mean Value Theorem. |
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| Example: |
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| Verify mean value theorem for the function |
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| f (x) = (x - 4) (x - 6) (x - 8) in [4,10] |
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| Step1: |
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| We know that every polynomial function is continuous and product of continues functions are continuous. f (x), being product of polynomials of degree 1, is a continuous function in [4,10]. |
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| Step 2: |
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| f ' (x) = (x - 6) (x - 4) + (x - 4) (x - 8) + (x - 6) (x - 8) |
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| f '(x)= (x2 -10x + 24) + (x2 - 12x + 32)+ (x2 - 14x + 48) |
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| = 3x2 - 36x + 104 |
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| f '(x) is defined for all values on the interval (4,10). |
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| \ f '(x) is differentiable. |
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| Step 3: |
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| Since both the condition are satisfied, Mean Value Theorem is applicable. |
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\ There exist c  (4, 10) such that |
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| \ f (4) = 0 |
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| f '(c) = 3c2 - 36c + 104 |
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| Substituting these values in (1), we have |
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Since 8  (4,10), Mean Value Theorem is satisfied. |
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