Let f(x)=x^3+2x^2-x be a real-valued function | vedclass

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(B) Given $f(x)=x^3+2x^2-x$. Since $f(x)$ is a polynomial,it is continuous on $[-1,2]$ and differentiable on $(-1,2)$. By Lagrange's Mean Value Theore

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$\frac{-2+\sqrt{19}}{3}$

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(B) Given $f(x)=x^3+2x^2-x$. Since $f(x)$ is a polynomial,it is continuous on $[-1,2]$ and differentiable on $(-1,2)$. By Lagrange's Mean Value Theorem,there exists $C \in (-1,2)$ such that $f'(C) = \frac{f(2)-f(-1)}{2-(-1)}$. First,calculate $f(2) = 2^3 + 2(2^2) - 2 = 8 + 8 - 2 = 14$. Next,calculate $f(-1) = (-1)^3 + 2(-1)^2 - (-1) = -1 + 2 + 1 = 2$. Now,$f'(x) = 3x^2 + 4x - 1$. So,$3C^2 + 4C - 1 = \frac{14-2}{3} = \frac{12}{3} = 4$. $3C^2 + 4C - 5 = 0$. Using the quadratic formula $C = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$,we get $C = \frac{-4 \pm \sqrt{16 - 4(3)(-5)}}{2(3)} = \frac{-4 \pm \sqrt{16+60}}{6} = \frac{-4 \pm \sqrt{76}}{6}$. Simplifying,$C = \frac{-4 \pm 2\sqrt{19}}{6} = \frac{-2 \pm \sqrt{19}}{3}$. Since $C \in (-1,2)$,we take the positive root: $C = \frac{-2+\sqrt{19}}{3}$.

Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then,the value of Lagrange's constant $C$ in $(-1,2)$ is

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