determine whether rolles theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that…

determine whether rolles theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)\n( f(x)=\frac{x^{2}-64}{x - 6}, quad-8,8 )\nyes, rolles theorem can be applied.\nno, because ( f ) is not continuous on the closed interval ( a, b ).\nno, because ( f ) is not differentiable on the open interval ( (a, b) ).\nno, because ( f(a) \neq f(b) ).\nif 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.)\n( c= )

determine whether rolles theorem can be applied to ( f ) on the closed interval ( a, b ). (select all that apply.)\n( f(x)=\frac{x^{2}-64}{x - 6}, quad-8,8 )\nyes, rolles theorem can be applied.\nno, because ( f ) is not continuous on the closed interval ( a, b ).\nno, because ( f ) is not differentiable on the open interval ( (a, b) ).\nno, because ( f(a) \neq f(b) ).\nif 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.)\n( c= )

Answer

Explanation:

Step1: Check continuity

A rational function (y = \frac{f(x)}{g(x)}) is continuous when (g(x)\neq0). 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\rightarrow c}f(x)=f(c)) for all (c\in[a,b])), since (x = 6\in[-8,8]) and (\lim_{x\rightarrow6}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)