ap calc review

Question 1

Extreme Value Theorem:
If f is continuous on the closed interval [a,b], then…
(yes calculator)

Answer 1

f attains an absolute maximum value and an absolute minimum value on [a,b].

Question 2

Does there exist a differentiable function f with f(0)= -1, f(2)=4, and f’(x) is < = 0 for all x? Sketch an example or explain why not.
(yes calculator)

Answer 2

no
the MVT guarantees the existence of a point c on the interval (0,2) with
f’(c) = [f(2)- f(0)]/2-0 = (4-(-1))/2 =5/2
So any differentiable function with f(0)= -1 and f(2) = 4 can’t possibly have f’(x) <= 0 for all x. no such f exists.