f(x) = x3 - 2x2 - x + 3
Since f(x) is polynomial function. So, f(x) is continuous in [0, 1] and differentiable in (0, 1).
Thus, both conditions of Lagrange's mean value theorem is appplicable.
Thus, there exist a point $\text{c}\in(0,1)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
$\Rightarrow3\text{c}^2-4\text{c}-1=\frac{[(1)^3-2(1)^2-(1)+3]-3}{1}$
⇒ 3c2 - 4c - 1 = 1-2
⇒ 3c2 - 4c + 1 = 0
⇒ 3c2 - 3c - c + 1 = 0
⇒ 3c(c - 1) - 1(c - 1) = 0
⇒ (3c - 1)(c - 1) = 0
$\Rightarrow\text{c}=\frac{1}{3}\in(0,1)$
Hence, Lagrange's mean value theorem is verified.
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