If f(x) = begin{cases} x^2-1, & 0

Question

(B) Given $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \ 2x+3, & 2 \leq x < 3 \end{cases}$.First,we find the left-hand limit at $x=2$:$\lim_{x ightarro

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$x^2-10x+21=0$

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(B) Given $f(x) = \begin{cases} x^2-1, & 0 First,we find the left-hand limit at $x=2$:$\lim_{x ightarrow 2^{-}} f(x) = \lim_{x ightarrow 2^{-}} (x^2-1) = 2^2 - 1 = 3$.Next,we find the right-hand limit at $x=2$:$\lim_{x ightarrow 2^{+}} f(x) = \lim_{x ightarrow 2^{+}} (2x+3) = 2(2) + 3 = 7$.The roots of the required quadratic equation are $\alpha = 3$ and $\beta = 7$.The quadratic equation is given by $x^2 - (\alpha + \beta)x + (\alpha 	imes \beta) = 0$.Substituting the values: $x^2 - (3 + 7)x + (3 	imes 7) = 0$.Thus,the equation is $x^2 - 10x + 21 = 0$.

If $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases}$,the quadratic equation whose roots are $\lim_{x \rightarrow 2^{-}} f(x)$ and $\lim_{x \rightarrow 2^{+}} f(x)$ is

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