(b) f ′ has a relative minimum at x = d, because f ′′ equals 0 at d, is less
than 0 on (b, d), and is greater than 0 on (d, g). Thus f ′ has a relative
minimum (from part a) at x = 2.651.
(c) The graph of f ′ has a point of inflection wherever its second
derivative f ′′′ changes from positive to negative or vice versa. This is
equivalent to f ′′ changing from increasing to decreasing (as at a and g) or
vice versa (as at c). Therefore, the graph of f ′ has three points of
inflection on [−1,3.2].
Graph f(x) = cos x and g(x) = x^2 − 1 in [−2,2] × [−2,2]. Here, y 1 = f and y 2
= g.
(a) Solve cos x = x^2 − 1 to find the two points of intersection:
(1.177,0.384) and (−1.177,0.384).
(b) Since ΔA = (y 1 − y 2 ) Δx = [f(x) − g(x)] Δx, the area A bounded by the
two curves is
(a) letters
(b) Draw a horizontal line at y = 20 (as shown on the graph below),
