How do you verify that the function #f(x)=x^3 - 21x^2 + 80x + 2# satisfies Rolle's Theorem on the given interval [0,16] and then find all numbers c that satisfy the conclusion of Rolle's Theorem?

Question

Cameron Alexander · Accepted Answer

Rolle's Theorem has three hypotheses: H1 :  #f# is continuous on the closed interval #[a,b]# H2 :  #f# is differentiable on the open interval #(a,b)#. H3 :  #f(a)=f(b)# In this question, #f(x)=x^3 - 21x^2 + 80x + 2# , #a=0# and #b=16#. We can apply Rolle's Theorem if all 3 hypotheses are true. So answer each question: H1 :  Is #f(x)=x^3 - 21x^2 + 80x + 2# continuous on the closed interval #[0,16]#? H2 :  Is #f(x)=x^3 - 21x^2 + 80x + 2#  differentiable on the open interval #(0,16)#? H3 :  Is #f(0)=f(16)#? If the answer to all three questions is yes, then Rolle's can be applied to this function on this interval. To solve #f'(c) = 0#, find #f'(x)#, set it equal to #0# and solve the equation.  (There may be more than one solution.) 
Select, as #c#, any solutions in #(0,16)#
(That is where Rolle's says there must be a solution.  There may be more than one solution in the interval.) Arithmetic note:
#f(16) = 16(16^2)-21(16^2)+80(16)+2# since #80 = 5*16#, we get:  #f(16) = 16(16^2)-21(16^2)+5*16 (16)+2#
so  #f(16) = 16(16^2)-21(16^2)+5*(16^2)+2 = 21(16^2)-21(16^2)+2=2# Please don't waste any of your youth finding #21*(16^2)#.)

Evelyn Agard · Answer

undefined To verify if $ f(x) = x^3 - 21x^2 + 80x + 2 $ satisfies Rolle's Theorem on the interval $[0, 16]$, first check if the function is continuous on the closed interval $[0, 16]$ and differentiable on the open interval $(0, 16)$. Then, check if $ f(0) = f(16) $. If these conditions are met, Rolle's Theorem states that there exists at least one number $ c $ in the open interval $(0, 16)$ such that $ f'(c) = 0 $.

1. Check continuity:
   $ f(x) $ is a polynomial function, so it is continuous everywhere.

2. Check differentiability:
   $ f(x) $ is a polynomial function, so it is differentiable everywhere.

3. Check if $ f(0) = f(16) $:
   $ f(0) = (0)^3 - 21(0)^2 + 80(0) + 2 = 2 $
   $ f(16) = (16)^3 - 21(16)^2 + 80(16) + 2 = 1634 $
   Since $ f(0) $ is not equal to $ f(16) $, proceed to find $ c $ that satisfies $ f'(c) = 0 $.

4. Find $ f'(x) $:
   $ f'(x) = 3x^2 - 42x + 80 $

5. Set $ f'(x) = 0 $ and solve for $ x $:
   $ 3x^2 - 42x + 80 = 0 $
   Using the quadratic formula: $ x = \frac{{42 \pm \sqrt{{(-42)^2 - 4 \cdot 3 \cdot 80}}}}{2 \cdot 3} $
   $ x = \frac{{42 \pm \sqrt{{-396}}}}{6} $
   $ x = \frac{{42 \pm 6i\sqrt{11}}}{6} $
   Since $ x $ must be a real number, the imaginary roots are discarded.
   $ x = \frac{{42 \pm 6\sqrt{11}}}{6} $
   $ x = 7 \pm \sqrt{11} $

Therefore, there are two values of $ c $ in the interval $(0, 16)$ that satisfy the conclusion of Rolle's Theorem: $ c = 7 + \sqrt{11} $ and $ c = 7 - \sqrt{11} $.