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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