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(C) First,analyze $g(t) = t^3 - 3t$. Its derivative $g'(t) = 3t^2 - 3 = 3(t-1)(t+1)$. Local maximum at $t = -1$ is $g(-1) = 2$. For $x \leq 2$,$f(x) =

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$(4, \frac{27}{4})$

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(C) First,analyze $g(t) = t^3 - 3t$. Its derivative $g'(t) = 3t^2 - 3 = 3(t-1)(t+1)$. Local maximum at $t = -1$ is $g(-1) = 2$. For $x \leq 2$,$f(x) = \max_{t \leq x} \{t^3 - 3t\}$. For $x \leq -1$,$f(x) = x^3 - 3x$. For $-1 For $2 For $3 \leq x For $4 \leq x For $x = 5$,$f(5) = [5-3] + 9 = 11$. For $x > 5$,$f(x) = 2x + 1$. Checking differentiability: At $x = -1$: $f(-1) = 2$,$f'(-1^-) = 0$,$f'(-1^+) = 0$. Differentiable. At $x = 2$: $f(2^-) = 2$,$f(2^+) = 2^2 + 2(2) - 6 = 2$. $f'(2^-) = 0$,$f'(2^+) = 2(2) + 2 = 6$. Not differentiable. At $x = 3$: $f(3^-) = 3^2 + 2(3) - 6 = 9$,$f(3^+) = 9$. $f'(3^-) = 2(3) + 2 = 8$,$f'(3^+) = 0$. Not differentiable. At $x = 4$: $f(4^-) = 9$,$f(4^+) = 10$. Discontinuous,so not differentiable. At $x = 5$: $f(5^-) = 10$,$f(5^+) = 11$. Discontinuous,so not differentiable. Thus,$m = 4$. $I = \int_{-2}^{-1} (x^3 - 3x) dx + \int_{-1}^{2} 2 dx = [\frac{x^4}{4} - \frac{3x^2}{2}]_{-2}^{-1} + [2x]_{-1}^{2} = ((\frac{1}{4} - \frac{3}{2}) - (4 - 6)) + (4 - (-2)) = (-\frac{5}{4} + 2) + 6 = \frac{3}{4} + 6 = \frac{27}{4}$. The ordered pair is $(4, \frac{27}{4})$.

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