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(B) Given $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$. Since $\frac{3}{2} > 1$,we use the definition $f(x) = cx^2 + bx + a$. $c(\frac{3}{2})^2 + b(\f

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$-8$

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(B) Given $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$. Since $\frac{3}{2} > 1$,we use the definition $f(x) = cx^2 + bx + a$. $c(\frac{3}{2})^2 + b(\frac{3}{2}) + a = 14 \Rightarrow \frac{9c}{4} + \frac{3b}{2} + a = 14 \Rightarrow 9c + 6b + 4a = 56$ ... $(i)$ Since $f(x)$ is continuous at $x = -1$,$\lim_{x \rightarrow -1^-} f(x) = \lim_{x \rightarrow -1^+} f(x) = f(-1)$. $a(-1)^2 + b(-1) + c = 2(-1)^2 + 4(-1) + 1 \Rightarrow a - b + c = -1$ ... (ii) Since $f(x)$ is continuous at $x = 1$,$\lim_{x \rightarrow 1^-} f(x) = \lim_{x \rightarrow 1^+} f(x) = f(1)$. $2(1)^2 + 4(1) + 1 = c(1)^2 + b(1) + a \Rightarrow a + b + c = 7$ ... (iii) Adding (ii) and (iii): $2a + 2c = 6 \Rightarrow a + c = 3 \Rightarrow c = 3 - a$. Subtracting (ii) from (iii): $2b = 8 \Rightarrow b = 4$. Substitute $b = 4$ and $c = 3 - a$ into $(i)$: $9(3 - a) + 6(4) + 4a = 56 \Rightarrow 27 - 9a + 24 + 4a = 56 \Rightarrow -5a = 5 \Rightarrow a = -1$. Then $c = 3 - (-1) = 4$. For $x = -2$,$f(x) = ax^2 + bx + c$. $\lim_{x \rightarrow -2} f(x) = a(-2)^2 + b(-2) + c = 4a - 2b + c = 4(-1) - 2(4) + 4 = -4 - 8 + 4 = -8$.

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

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