1. Locate the absolute extrema of the function f(x)=cos(pi*x) on the closed interval [0,1/2].
2. Determine whether Rolle's Theorem applied to the function f(x)=x^2+6x+8 on the closed interval[-4,-2]. If Rolle's Theorem can be applied, find all values of c in the open interval (-4,-2) such that f'(c)=0.
3. Determine whether the open intervals on which the graph of f(x)=-7x+7cosx is concave upward or downward.
4. Find the points of inflection and discuss the concavity of the function f(x)=sinx-cosx on the interval (0,2pi)
5.Find the points of inflection and discuss the concavity of the function f(x)=-x^3+x^2-6x-5
Calculus - Steve, Wednesday, October 31, 2012 at 10:38am
f' = -pi sin(pi*x) extrema where f' = 0, or x an integer
since f(x) = (x+4)(x+2) f(-4)=f(-2)=0, so we're good to do. vertex is at x = -3.
f is concave up if f'' > 0
f'' = -7cosx, so where is that >0? <0?
concavity as above, inflection where f'' = 0
f'' = -sinx + cosx = √2 sin(x + π/4)
same methods as in #3,4/
f'' = -6x