Given the function g[:1:) = 42:3 301:2 + 4822, find the first derivative, g'(:::). 9'03) = Notice that g'(:r:) = Owhen a: = 1, that is, g'(1) = 0. Now, we want to know whether there is a local minimum or local maximum at a: = 1, so we will use the second derivative test. Find the second derivative, 9' (3:) 9' '(m) = i i Evaluate g' '(1). 9' '(1) = [:1 Based on the sign of this number, does this mean the graph of g(:c) is concave up or concave down at a: = 1? [Answer either up or down -- watch your spelling!!] At a: = 1 the graph of g(:t:) is concave Based on the concavity of 9(3) at a: = 1, does this mean that there is a local minimum or local maximum at :1: = 1? [Answer either minimum or maximum -- watch your spelling!!]