How do you verify that the function f(x)=x^3 - 21x^2 + 80x + 2f(x)=x321x2+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?

1 Answer
May 26, 2015

Rolle's Theorem has three hypotheses:

H1 : ff is continuous on the closed interval [a,b][a,b]

H2 : ff is differentiable on the open interval (a,b)(a,b).

H3 : f(a)=f(b)f(a)=f(b)

In this question, f(x)=x^3 - 21x^2 + 80x + 2f(x)=x321x2+80x+2 , a=0a=0 and b=16b=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 + 2f(x)=x321x2+80x+2 continuous on the closed interval [0,16][0,16]?

H2 : Is f(x)=x^3 - 21x^2 + 80x + 2f(x)=x321x2+80x+2 differentiable on the open interval (0,16)(0,16)?

H3 : Is f(0)=f(16)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).)