Average rate of change is the same as slope through the two endpoints of the interval
For f(x), (1,-2) and (3,6) are on the graph
Slope m = (y2-y1)/(x2-x1) m = (6-(-2))/(3-1) m = (6+2)/(3-1) m = 8/2 m = 4 So we can rule out f(x) because we want the slope to be 3, not 4.
------------------------- Let's try g(x)
Based on the given curve, we can see that (1,-2) and (3,4) are on the g(x) graph.
Slope m = (y2-y1)/(x2-x1) m = (4-(-2))/(3-1) m = (4+2)/(3-1) m = 6/2 m = 3 We found the answer, but let's check h(x) just for the sake of completeness -------------------------
If we plug x = 1 and x = 3 into h(x), we get y = 1/2 and y = 19/8 respectively So the two points (1,1/2) and (3,19/8) are on the graph
Slope m = (y2-y1)/(x2-x1) m = (19/8-1/2)/(3-1) m = (19/8-4/8)/(3-1) m = (15/8)/(2) m = (15/8)*(1/2) m = 15/16 which is not equal to 3, so we can rule out h(x)
