Question

determine whether rolles theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)
( f(x)=\frac{x^{2}-64}{x - 6}, quad-8,8 )
yes, rolles theorem can be applied.
no, because ( f ) is not continuous on the closed interval ( a, b ).
no, because ( f ) is not differentiable on the open interval ( (a, b) ).
no, because ( f(a)
eq f(b) ).
if rolles theorem can be applied, find all values of ( c ) in the open interval ( (a, b) ) such that ( f^{prime}(c)=0 ). (enter your answers as a comma-separated list. if rolles theorem cannot be applied, enter na.)
( c= )

Explanation:

Step1: Check continuity

A rational function \(y = \frac{f(x)}{g(x)}\) is continuous when \(g(x)
eq0\). For \(f(x)=\frac{x^{2}-64}{x - 6}=\frac{(x + 8)(x - 8)}{x - 6}\), the function is not continuous at \(x = 6\) which lies in the interval \([-8,8]\). By the definition of continuity (a function \(y = f(x)\) is continuous on \([a,b]\) if \(\lim_{x
ightarrow c}f(x)=f(c)\) for all \(c\in[a,b]\)), since \(x = 6\in[-8,8]\) and \(\lim_{x
ightarrow6}f(x)\) does not equal \(f(6)\) (because \(f(6)\) is undefined), \(f(x)\) is not continuous on \([-8,8]\).

Step2: Check differentiability (implied by non - continuity)

Since a function must be continuous on \([a,b]\) to be differentiable on \((a,b)\) (by the relationship between continuity and differentiability: if \(y = f(x)\) is differentiable at \(x=c\), then \(y = f(x)\) is continuous at \(x = c\)), and \(f(x)\) is not continuous on \([-8,8]\), it is also not differentiable on \((-8,8)\) in the context of Rolle's Theorem requirements.

Step3: Check \(f(a)=f(b)\) (though non - continuity already makes the theorem inapplicable)

Calculate \(f(-8)=\frac{(-8)^{2}-64}{-8 - 6}=\frac{64 - 64}{-14}=0\) and \(f(8)=\frac{8^{2}-64}{8 - 6}=\frac{64 - 64}{2}=0\). But the non - continuity condition already violates Rolle's Theorem.

Answer:

No, because \(f\) is not continuous on the closed interval \([a,b]\); No, because \(f\) is not differentiable on the open interval \((a,b)\); \(NA\)