Let f(x)=(x-1)(x-2)(x-3),where x in [0,4]. Fi | vedclass

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. Since $f(x)$ is a polynomial,it is continuous on $[0, 4]$ and differentiable on $(0, 4)$. B

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$\frac{6-2 \sqrt{3}}{3}, \frac{6+2 \sqrt{3}}{3}$

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(B) Given $f(x) = (x-1)(x-2)(x-3) = x^3 - 6x^2 + 11x - 6$. Since $f(x)$ is a polynomial,it is continuous on $[0, 4]$ and differentiable on $(0, 4)$. By $LMVT$,there exists at least one $c \in (0, 4)$ such that $f'(c) = \frac{f(4) - f(0)}{4 - 0}$. First,calculate $f(4) = (4-1)(4-2)(4-3) = 3 	imes 2 	imes 1 = 6$. Next,calculate $f(0) = (0-1)(0-2)(0-3) = -1 	imes -2 	imes -3 = -6$. So,$f'(c) = \frac{6 - (-6)}{4} = \frac{12}{4} = 3$. Now,$f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 11x - 6) = 3x^2 - 12x + 11$. Setting $f'(c) = 3$,we get $3c^2 - 12c + 11 = 3$,which simplifies to $3c^2 - 12c + 8 = 0$. Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $c = \frac{12 \pm \sqrt{144 - 4(3)(8)}}{2(3)} = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm \sqrt{48}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = \frac{6 \pm 2\sqrt{3}}{3}$.

Let $f(x)=(x-1)(x-2)(x-3)$,where $x \in [0,4]$. Find the values of $c$ if Lagrange's Mean Value Theorem $(LMVT)$ can be applied.

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