Verify Lagranges mean value theorem for function `f(x)=sqrt(x^2-4)` on `[2,\ 4]` and find a point `' c '` in the indicated interval:

Answer

`c=frac{3}{2}, frac{-1}{2}`

Explanation

Given,
`f(x)=\sqrt{x^{2}-x}, x \in[1,4]`


As `x^{2}-x` is continuous on `R`


and `\sqrt{x^{2}-x}` is exists in `[1,4]`


`\Rightarrow f(x)` is continuous in `[1,4]`.



Differentiating the given function w.r.t `x`


`f^{\prime}(x)=\frac{1}{2}(x^{2}-x)^{-1 / 2} \cdot(2 x-1)=\frac{2 x-1}{2 \sqrt{x^{2}-x}}`


which exists `\forall x \in R`.



`\therefore f(x)` is differentiable in `(1,4)`.



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