1. ## Rolle's Theorem

State why Rolle's Theorem does not apply to the function f(x) = $\displaystyle \frac {2} {(x+1)^2}$ on the interval [-2, 0]

a) f is not continuous on [-2,0]
b) f(-2) not equal f(0)
c) f is not differentiable at x = -1
d) Both a and c
e) None of these

2. What is the domain of the function f ?

3. Originally Posted by Samantha
State why Rolle's Theorem does not apply to the function f(x) = $\displaystyle \frac {2} {(x+1)^2}$ on the interval [-2, 0]

a) f is not continuous on [-2,0]
b) f(-2) not equal f(0)
c) f is not differentiable at x = -1
d) Both a and c
e) None of these
Rolle's theorem doesn't apply because the function is not continuous on [-2, 0]. (There is a discontinuity at x = -1.)

a) I already said why this is true.

b) f(-2) does equal f(0).
$\displaystyle f(-2) = \frac{2}{(-2 + 1)^2} = \frac{2}{(-1)^2} = \frac{2}{1} = 2$
and
$\displaystyle f(0) = \frac{2}{(0 + 1)^2} = \frac{2}{(1)^2} = \frac{2}{1} = 2$

c) f(x) is, in fact, not differentiable at x = -1 because the function does not exist there.

As both a) and c) are true, and Rolle's theorem depends on them being true (and f(-2) = f(0) to boot) the answer should be d).

-Dan