Given f'(x)=(x-4)(6-2x) find the x-coordinate for the relative minimum of the graph f(x).
options:
8
6
3
none of these
i think it is 3, but i am also conflicted to say none of these because i graphed the function and ant see a minimum.

The graph you plotted is the graph of f ' (x) and NOT f(x) itself. 

Draw a number line. On the number line plot x = 3 and x = 4. These values make f ' (x) equal to zero. Pick a value to the left of x = 3, say x = 0. Plug in x = 0 into the derivative function to get

f ' (x) = (x-4)(6-2x)
f ' (0) = (0-4)(6-2*0)
f ' (0) = -24

So the function is decreasing on the interval to the left of x = 3. Now plug in a value between 3 and 4, say x = 3.5

f ' (x) = (x-4)(6-2x)
f ' (3.5) = (3.5-4)(6-2*3.5)
f ' (3.5) = 0.5

The function is increasing on the interval 3 < x < 4. The junction where it changes from decreasing to increasing is at x = 3. This is where the min happens.

So the final answer is C) 3