Find a cubic function that has horizontal tangents at two given points

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To find a cubic function with horizontal tangents at the points (-2, 6) and (2, 0), the first derivative must equal zero at these x-values. This leads to two equations from the derivative, which can be combined with the original function evaluated at the given points to form a system of four equations with four unknowns (a, b, c, d). The discussion highlights the importance of correctly setting up these equations and solving them systematically, either through substitution or matrix methods. An error in the setup of the equations can lead to incorrect results, emphasizing the need for careful verification of each step. Ultimately, the correct cubic function can be derived by ensuring all equations are compatible and accurately represent the conditions given.
  • #31
As long as I have been working problems like this, and teaching Linear Algebra (one of my favorite courses) I still consider writing a system of equations as a matrix and row reducing to be more trouble than it is worth!

A general cubic is ax^3+ bx^2+ cx+ d. The fact that the point (2, 0) is on graph means that 8a+ 4b+ 2c+ d= 0. The fact that (-2, 6) is on the graph means that -8a+ 4b- 2c+ d= 6. The fact that there is a horizontal tangent at x= 2 means that 12a+ 4b+ c= 0. The fact that there is a horizontal tangent at x= -2 means that 12a- 4b+ c= 0. We need to solve those 4 equations for a, b, c, and d.

Adding 8a+ 4b+ 2c+ d= 0 and -8a+ 4b-2c+ d= 6 gives 8b+ 2d= 6 or d= 3- 4b. Adding 12a+ 4b+ c= 0 and 12a- 4b+ c= 0 gives 24a+ 2c= 0 or c= -12a. Putting those into the first and third equations gives 8a+ 4b+ 2(-12a)+ 3-4b= -16a+ 3= 0 or a= 3/16 and 12a+ 4b+ (-12a)= 4b= 0. Then c= -12a= -36/16= -9/4 and d= 3.
 

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