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# So i found that since at x =2 of the graph of g is undefined, although on the derivative x=2 would make it undefined (a critical value), it wouldnt...

So i found that since at x =2 of the graph of g is undefined, although on the derivative x=2 would make it undefined (a critical value), it wouldnt be because x=2 is undefined on the original graph. WIth that said, the only critical points are -4 and 4. However, for part B to identify local mins or max, I did a sign diagram for -4 and 4 and found that at x=4, to the left and right of the sign diagram is positive so that would make it neither a min or max. However, when i integrated the derivative function and plugged in x=3 and y=4 to get a C and graphed, I saw that 4 appears to be a minimum and when I did the second derivative and plugged in 4 for x i got a positive value which makes it a min. So my answers are conflicting!