Answer:
Local minimum at x = 0.
Step-by-step explanation:
Local minimums occur when g'(x) = 0 and g"(x) > 0.
Local maximums occur when g'(x) = 0 and g"(x) < 0.
Set g'(x) equal to 0 and solve:
0 = 2x (x − 1)² (x + 1)²
x = 0, 1, or -1
Evaluate g"(x) at each point:
g"(0) = 2
g"(1) = 0
g"(-1) = 0
There is a local minimum at x = 0.
V = S*h
h = V/S = 250*pi / 50*pi = 5 m
A) (f + g)(x) = f(x) + g(x) = (5x-12) + (11x+2) = 16x-10
B) (f × g)(x) = f(x)×g(x) = (5x-12)×(11x+2) = 5x(11x) -12(11x)+2(5x) -12(2) = 55x² -122x -24
C) f(g(x)) = 5(11x+2) -12 . . . . substitute g(x) for the x in f(x)
... = 55x +10 -12 = 55x -2
11. C
12. D
13. x = 1.35
14. A
