For the function `f(x)=x+1/x ,\ \ x in [1,\ 3]` , the value of `c` for the Lagranges mean value theorem is (a) 1 (b) `sqrt(3)` (c) 2 (d) none of these

Answer

`C=sqrt{3}`

Explanation

Given,

`f(x)=x+\frac{1}{x}=\frac{x^{2}+1}{x}`


We can see, `f(x)` is continuous on

`[1,3]` and derivable on `(1,3)`.

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there existsc `\in(1,3)` such that

`f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}`


Now, `f(x)=\frac{x^{2}+1}{x}`


`f^{\prime}(x)=\frac{x^{2}-1}{x^{2}}, f(1)=2, f(3)=\frac{10}{3}`


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