ln(x+2) is defined and differentiable on (-2,∞) so (a) is out
Rolle's Theorem is a special case of the Mean Value Theorem, where f(a) = f(b)
Clearly, that is not the case here...
Rolle's theorem cannot be applied t the function f(x)= ln(x+2) on the interval [-1,2] because
a) f is not differentiable on the interval [-1,2]
b) f(-1)≠ f(2)
c) All of these
d) Rolle's theorem can be applied to f(x)= ln(x+2) on the interval [-1,2]
2 answers
ngg