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

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$\frac{3}{2}$ We have f (x) = x (x − 2) It can be rewritten as f\left( x \right) = x^2 - 2x . We know that a polynomial function is everywhere continuous and differentiable. Since f\left( x \right) is a polynomial , it is continuous on \left[ 1, 2 \right] and differentiable on \left( 1, 2 \right) . Thus, f\left( x \right) satisfies both the conditions of Lagrange's theorem on \left[ 1, 2 \right] . So, there must exist at least one real number c \left[ 1, 2 \right] such that f'\left( c \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1} = \frac{f\left( 2 \right) - f\left( 1 \right)}{1} Now, f\left( x \right) = x^2 - 2x \Rightarrow f'\left( x \right) = 2x - 2, andf\left( 1 \right) = - 1, f\left( 2 \right) = 0 \therefore f'\left( x \right) = \frac{f\left( 2 \right) - f\left( 1 \right)}{2 - 1} \Rightarrow f'\left( x \right) = 0 + $\frac{1}{1}$ \Rightarrow 2x - 2 = 1 \Rightarrow x = $\frac{3}{2}$ $\therefore$ c = $\frac{3}{2}$ \in \left( 1, 2 \right).

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