If f(x) = begin{cases} 1 + cos x, & x le 0 a | vedclass

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(B) Given that $f(x)$ is continuous at $x=0$ and $x=2$. For continuity at $x=0$, we must have $f(0^-) = f(0) = f(0^+)$. $f(0^-) = \lim_{x 	o 0^-} (1

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$8$

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(B) Given that $f(x)$ is continuous at $x=0$ and $x=2$. For continuity at $x=0$, we must have $f(0^-) = f(0) = f(0^+)$. $f(0^-) = \lim_{x 	o 0^-} (1 + \cos x) = 1 + \cos(0) = 2$. $f(0^+) = \lim_{x 	o 0^+} (a - x) = a - 0 = a$. Equating these, we get $a = 2$. For continuity at $x=2$, we must have $f(2^-) = f(2) = f(2^+)$. $f(2^-) = \lim_{x 	o 2^-} (a - x) = a - 2 = 2 - 2 = 0$. $f(2^+) = \lim_{x 	o 2^+} (x - b)^2 = (2 - b)^2$. Equating these, we get $(2 - b)^2 = 0$, which implies $b = 2$. Finally, $a^2 + b^2 = (2)^2 + (2)^2 = 4 + 4 = 8$.

If $f(x) = \begin{cases} 1 + \cos x, & x \le 0 \\ a - x, & 0 < x < 2 \\ (x - b)^2, & x \ge 2 \end{cases}$ is continuous at $x=0$ and $x=2$, then find the value of $a^2+b^2$.

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