Rolle's Theorem?

Question

Find the number c that satisfies the conclusion of Rolle's Theorem.
f(x) = x3 - x2 - 20x + 3
[0, 5]

Answer 1

Roole's Theorem says that if f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b) = 0, then there exists a c in (a,b) such that f'(c) = 0.

Your function is a polynomial of degree 2, it's derivative is f'(x) = 3x^2 - 2x - 20, which has 2 real roots. So, since f has degree 2, f has has 3 real roots. There are 2 numbers c that satisfy Rolle's Theorem, they are the roots of f', easily computed by Bhaskara formula.

Answer 2

c = the positive root [2+sqrt 244] / 6

c simplifies to [ 1 + sqrt 61] / 3 = about 2.9367

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Rolle's theorem is a special case of (I can't remember the name) another theorem -- for a continuous function over the interval [a,b] there exists a "c" , a f'(c) = [ f(b) - f(a) ] / [b-a]

Answer 3

f(b)=f(a) there exist f'(c)=0 where c is inbetween [a,b]
...
f'(c)=3x^2-2x-20=0
solving with quadratic formula x = 2.93 or -2.26
since 2.93 is with [a,b], c=2.93