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)=(x - 5)(x + 4)^{2}, quad-4,5 )\nyes, rolles theorem can be applied.\nno, because ( f ) is not continuous on the closed interval ( a, b ).\nno, because ( f ) is not differentiable in 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)=(x - 5)(x + 4)^{2}, quad-4,5 )\nyes, rolles theorem can be applied.\nno, because ( f ) is not continuous on the closed interval ( a, b ).\nno, because ( f ) is not differentiable in 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 polynomial function (y=(x - 5)(x + 4)^{2}=x^{3}+3x^{2}-24x - 80) is continuous everywhere. So (f(x)) is continuous on ([-4,5]).

Step2: Check differentiability

The derivative of (y=(x - 5)(x + 4)^{2}) using the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u=x - 5), (u^\prime=1) and (v=(x + 4)^{2}), (v^\prime = 2(x + 4)). Then (f^\prime(x)=(x + 4)^{2}+2(x - 5)(x + 4)=(x + 4)(x + 4+2x-10)=(x + 4)(3x - 6)). A polynomial function is differentiable everywhere. So (f(x)) is differentiable on ((-4,5)).

Step3: Check (f(a)=f(b))

(f(-4)=(-4 - 5)(-4 + 4)^{2}=0) and (f(5)=(5 - 5)(5 + 4)^{2}=0). So (f(-4)=f(5)). Since (f(x)) is continuous on ([-4,5]), differentiable on ((-4,5)) and (f(-4)=f(5)), Rolle's Theorem can be applied.

Step4: Find (c)

Set (f^\prime(c)=0). Since (f^\prime(x)=(x + 4)(3x - 6)), then ((c + 4)(3c - 6)=0). Solving (c+4 = 0) gives (c=-4) (not in the open interval ((-4,5))), solving (3c-6=0) gives (c = 2).

Answer:

Yes, Rolle's Theorem can be applied. (c = 2)