If f(x) = begin{cases} x^2 - 3, & 2

Question

(C) Given $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \ 2x + 5, & 3 < x < 4 \end{cases}$.First,we find the left-hand limit as $x 	o 3^-$:$\lim_{x 	o

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$x^2 - 17x + 66 = 0$

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(C) Given $f(x) = \begin{cases} x^2 - 3, & 2 First,we find the left-hand limit as $x 	o 3^-$:$\lim_{x 	o 3^-} f(x) = \lim_{x 	o 3^-} (x^2 - 3) = 3^2 - 3 = 9 - 3 = 6$.Next,we find the right-hand limit as $x 	o 3^+$:$\lim_{x 	o 3^+} f(x) = \lim_{x 	o 3^+} (2x + 5) = 2(3) + 5 = 6 + 5 = 11$.The roots of the required quadratic equation are $6$ and $11$.The equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Sum of roots $= 6 + 11 = 17$.Product of roots $= 6 	imes 11 = 66$.Therefore,the equation is $x^2 - 17x + 66 = 0$.

If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

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