If f(x) = begin{cases} frac{x^2}{a} - a, & x | vedclass

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(B) Given $f(x) = \begin{cases} \frac{x^2}{a} - a, & x  a \end{cases}$.First,we find the value of the fun

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$f(x)$ is continuous at $x = a$

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(B) Given $f(x) = \begin{cases} \frac{x^2}{a} - a, & x  a \end{cases}$.First,we find the value of the function at $x = a$: $f(a) = 0$.Next,we calculate the left-hand limit $(LHL)$ at $x = a$:$\lim_{x 	o a^-} f(x) = \lim_{h 	o 0} f(a - h) = \lim_{h 	o 0} \left( \frac{(a - h)^2}{a} - a ight) = \frac{a^2}{a} - a = a - a = 0$.Then,we calculate the right-hand limit $(RHL)$ at $x = a$:$\lim_{x 	o a^+} f(x) = \lim_{h 	o 0} f(a + h) = \lim_{h 	o 0} \left( a - \frac{(a + h)^2}{a} ight) = a - \frac{a^2}{a} = a - a = 0$.Since $\lim_{x 	o a^-} f(x) = \lim_{x 	o a^+} f(x) = f(a) = 0$,the function $f(x)$ is continuous at $x = a$.

If $f(x) = \begin{cases} \frac{x^2}{a} - a, & x < a \\ 0, & x = a \\ a - \frac{x^2}{a}, & x > a \end{cases}$ then:

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