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Verify Lagrange’s mean value theorem for the function f(x)=x+1/x, x ∈ [1, 3]
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Solution
step 1:
Given
`f(x)=x+1/x, x ∈ [1, 3]`
We know that a polynomial function is continuous everywhere and also differentiable.
So f(x)" data-mce-style="display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 13px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;" data-mce-tabindex="0">f(x)f(x) being a polynomial is continuous and differentiable on (1,3)
So there must exist at least one real number c ∈ [1, 3] such that
`f'(c)=(f(3)-f(1))/(3-1)`
step2:
`f(x)=x+1/x`
`f(3)=10/3`
`f(1)=2`
`f'(x)=1-1/x^2`
`f'(c)=1-1/c^2`
`1-1/c^2=(10/3-2)/2`
`(c^2-1)/c^2=2/3`
`3c^2-3=2c^2`
`c^2=3`
`c=sqrt3`
`c in (1,3)`
Hence Lagrange's Mean Value theorem is verified
Concept: Mean Value Theorem
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