g is increasing when g' > 0
g'(x) = f(x)
f(x) > 0 when -3 < x < 6
g(x) = ∫[-2,x] 3 dt = 3t [-2,x] = -6-3x for -3<=x<0
g(x) = ∫[-2,x] 3-t dt = 3t - t^2/2 [-2,x] = x^2/2 - 3x - 8 for 0<=x<=6
g(x) = ∫[-2,x] -3 dt = -3t [-2,x] = 6-3x for 6<x<=9
Use the information in the image: imgur(.)com/a/YzAGLvs (Remove the parenthesis to use link)
1. On what interval is g increasing? Justify your answer.
2. For 0 ≤ x ≤ 6, express g(x) in terms of x. Do not include +C in your final answer.
1 answer