If the function f(x) = begin{cases} 1+cos x, | vedclass

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(B) For the function $f(x)$ to be continuous everywhere,it must be continuous at $x=0$ and $x=2$. At $x=0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-}

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$8$

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(B) For the function $f(x)$ to be continuous everywhere,it must be continuous at $x=0$ and $x=2$. At $x=0$: $\lim_{x 	o 0^-} f(x) = \lim_{x 	o 0^-} (1+\cos x) = 1+1 = 2$. $\lim_{x 	o 0^+} f(x) = \lim_{x 	o 0^+} (a-x) = a-0 = a$. Since the function is continuous at $x=0$,$a = 2$. At $x=2$: $\lim_{x 	o 2^-} f(x) = \lim_{x 	o 2^-} (a-x) = a-2 = 2-2 = 0$. $\lim_{x 	o 2^+} f(x) = \lim_{x 	o 2^+} (x^2-b^2) = 4-b^2$. Since the function is continuous at $x=2$,$0 = 4-b^2$,which implies $b^2 = 4$. Therefore,$a^2+b^2 = 2^2 + 4 = 4+4 = 8$.

If the function $f(x) = \begin{cases} 1+\cos x, & x \leq 0 \\ a-x, & 0 < x \leq 2 \\ x^2-b^2, & x > 2 \end{cases}$ is continuous everywhere,then $a^2+b^2=$

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