I have a line segment $ overline {AB} $with endpoints $ (x_1, y_1) $ and $ (x_2, y_2) $, Draw any function $ f (x) $Through these 2 points, the function must have an average speed $[x_1,x_2]$ from $ frac {y_2-y_1} {x_2-x_1} $, However, the area is bounded by the line and $ f (x) $ can be very different, and what I'm trying to find is the relationship between $ f & # 39; (x) $ and this area. I am not sure how to formally express this next part, but as the average speed of both the line and the function over $[x_1,x_2]$ is the same, the average of the sum of the values of $ f & # 39; (x) $ at all infinite points $ f (x) $ about the domain $[x_1,x_2]$ is the average of the sum of the slope / value of the derivative of the line at all infinite points of the line in the domain $[x_1,x_2]$, The actual values of $ f & # 39; (x) $ can vary considerably. Visually, it seems to be the larger y-value of $ f & # 39; (x) $ over $[x_1,x_2]$The bigger the area. For example, with the line segment with endpoints (0,0) $ and (2.4) $and the function $ f (x) = x ^ frac {3} {2} * sqrt {2} $the area is limited by $ y = 2x $ and $ f (x) $ is $ frac {16-2 sqrt {2}} {5} $and the largest y-value of $ f & # 39; (x) $ over $[1,2]$ is 3. If $ f (x) = frac {x ^ 3} {2} $is then the area 4 and the largest y-value of $ f & # 39; (x) $ over $[0,2]$ is 6.

Am I correct in saying the larger the largest y-value of is $ f & # 39; (x) $ over $[x_1,x_2]$The bigger the area? And if so, there is a relationship where a line segment is over $[x_1,x_2]$, can give the area bounded by the line segment and $ f (x) $, only the largest y-value of $ f & # 39; (x) $ over $[x_1,x_2]$?