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 - 1)(x - 4)(x - 5),quad1,5 )\n( square ) yes, rolles theorem can be applied.\n( square ) no, because ( f ) is not continuous on the closed interval (a,b).\n( square ) no, because ( f ) is not differentiable on the open interval ( (a,b) ).\n( square ) no, 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
Since (f(x)=(x - 1)(x - 4)(x - 5)) is a polynomial, it is continuous on the closed interval ([1,5]).
Step2: Check differentiability
Since (f(x)) is a polynomial, it is differentiable on the open interval ((1,5)).
Step3: Check (f(a)=f(b))
(f(1)=(1 - 1)(1 - 4)(1 - 5)=0) and (f(5)=(5 - 1)(5 - 4)(5 - 5)=0), so (f(1)=f(5)).
Step4: Differentiate (f(x))
First, expand (f(x)=(x - 1)(x^{2}-9x + 20)=x^{3}-9x^{2}+20x-x^{2}+9x - 20=x^{3}-10x^{2}+29x - 20). Then (f^{\prime}(x)=3x^{2}-20x + 29).
Step5: Solve (f^{\prime}(c)=0)
Set (3c^{2}-20c + 29 = 0). Using the quadratic formula (c=\frac{20\pm\sqrt{400-348}}{6}=\frac{20\pm\sqrt{52}}{6}=\frac{20\pm2\sqrt{13}}{6}=\frac{10\pm\sqrt{13}}{3}).
Answer:
Yes, Rolle's Theorem can be applied. (c=\frac{10+\sqrt{13}}{3},\frac{10 - \sqrt{13}}{3})