Question

determine whether rolles theorem can be applied to ( f ) on the closed interval (a,b). (select all that apply.)
( f(x)=(x - 1)(x - 4)(x - 5),quad1,5 )
( square ) yes, rolles theorem can be applied.
( square ) no, because ( f ) is not continuous on the closed interval (a,b).
( square ) no, because ( f ) is not differentiable on the open interval ( (a,b) ).
( square ) 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

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. \(c=\frac{10+\sqrt{13}}{3},\frac{10 - \sqrt{13}}{3}\)