The value of c in Largrange's mean value theorem for the function `f(x)=x(x-2)` when ` x in [1,2]` is

Answer

`c=frac{3}{2} in(1,2)`

Explanation

ATQ,
`f(x)=x(x-2)`


It can be rewritten as

`f(x)=x^{2}-2 x`.


We know that a polynomial function is continuous and differentiable at every point.

Since `f(x)` is a polynomial, it is continuous on `[1,2]` and differentiable on `(1,2)`.

Thus, `f(x)` satisfies the conditions of Lagrange's theorem on `[1,2]` range.

So, there must exist one real number `c [1,2]` such that

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