David R. answered • 06/24/20
Math, Statistics and History tutor (up through AP Calculus AB)
Hi Leona, I just want to confirm that the points you listed were for f ' (the derivative). If that is the case, then as long as the graph of the derivative has a y-value less than zero (position below the x-axis on the graph) then the original graph will have a negative slope. It seems if you connect the points, the graph of the derivative stays below the x-axis from x=-2 to x=2 (where there is a horizontal tangent, meaning the graph levels off flat), then continues back down into the area under the x-axis (until the next horizontal tangent at x=4) where the graph of the derivative will begin to rise again. The derivative never actually turns positive until after it crosses the x-axis at x=5 (point 5,0) as seen by the final point (6,4)... so what does all this mean?
The original function graph (f) is sloping downwards or at best leveling off from x=-2 to x=5, that means the absolute minimum will occur at x=5, the lowest point, before the slope (derivative) turns positive.
When checking for absolute minimums using the derivative function, you are looking at the starting and ending points, or any point in between where the slope turns from negative to positive. The starting point can't be a minimum because the slope coming off it is negative and the end point can't be a minimum because the slope at that point is positive. Somewhere in between the slope changed from negative to positive ===> (5,0) is the culprit where the derivative crosses zero (the x-axis) from negative to positive.
Tip: always try to sketch out the points given to you, then you can use that sketch to help you understand what is going on with other graphs f, f ', f "
I hope that helped, and God bless you on your studies.
