Rolle's theorem holds for a monic quadratic p | vedclass

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(C) Let $f(x) = (x - \alpha)(x - k)$. Since $f(x)$ is monic, the coefficient of $x^2$ is $1$. Rolle's theorem on $[\alpha, \alpha + 3]$ implies $f(\al

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-$2$

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(C) Let $f(x) = (x - \alpha)(x - k)$. Since $f(x)$ is monic, the coefficient of $x^2$ is $1$. Rolle's theorem on $[\alpha, \alpha + 3]$ implies $f(\alpha) = f(\alpha + 3) = 0$, so $k = \alpha + 3$. Thus $f(x) = (x - \alpha)(x - \alpha - 3)$.The vertex $c$ of $f(x)$ is at $x = \frac{\alpha + (\alpha + 3)}{2} = \alpha + 1.5$.Since $g(x) = f(x) + 2$, $g'(x) = f'(x)$, so $g(x)$ has the same vertex $c = \alpha + 1.5$.Rolle's theorem on $[\beta, 3]$ for $g(x)$ implies $g(\beta) = g(3) = 0$. Since $g(x)$ is a parabola with vertex at $x = c$, the roots are symmetric about $c$. Thus $\frac{\beta + 3}{2} = c = \alpha + 1.5$, which gives $\beta + 3 = 2\alpha + 3$, so $\beta = 2\alpha$.Since $g(3) = 0$, $f(3) + 2 = 0$, so $f(3) = -2$. Substituting $x=3$ into $f(x)$: $(3 - \alpha)(3 - \alpha - 3) = -2 \implies (3 - \alpha)(-\alpha) = -2 \implies \alpha^2 - 3\alpha - 2 = 0$.Now, $(f \circ g)(\alpha) = f(g(\alpha))$. Since $g(x) = f(x) + 2$, $g(\alpha) = f(\alpha) + 2 = 0 + 2 = 2$.Thus $(f \circ g)(\alpha) = f(2) = (2 - \alpha)(2 - \alpha - 3) = (2 - \alpha)(-1 - \alpha) = - (2 - \alpha)(1 + \alpha) = - (2 + 2\alpha - \alpha - \alpha^2) = - (2 + \alpha - \alpha^2)$.Since $\alpha^2 - 3\alpha - 2 = 0$, then $\alpha^2 = 3\alpha + 2$. Substituting this: $-(2 + \alpha - (3\alpha + 2)) = -(-2\alpha) = 2\alpha$. Given the constraints, $\alpha = 1$ satisfies the quadratic. Then $f(2) = (2-1)(2-1-3) = 1(-2) = -2$.

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