What Happens at the Cusp on the Original Graph When Viewing the Derivative?

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SUMMARY

The discussion centers on the behavior of derivatives at cusps in original graphs. When the original function, f, has a cusp, the derivative f' is undefined at that point, leading to an asymptote. Conversely, if the derivative graph f' exhibits a cusp, it indicates that the second derivative does not exist. The example of y = |x| illustrates this concept, where f'(x) = |x| leads to f(x) = (1/2)x² for x ≥ 0 and -(1/2)x² for x < 0, demonstrating the cusp's impact on the original graph.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and integrals.
  • Familiarity with the properties of cusps in graph analysis.
  • Knowledge of piecewise functions and their graphical representations.
  • Basic comprehension of second derivatives and their significance.
NEXT STEPS
  • Study the implications of cusps in calculus, focusing on differentiability.
  • Explore the concept of second derivatives and their role in determining concavity.
  • Investigate piecewise function integration techniques and their applications.
  • Learn about the graphical interpretation of derivatives and their continuity.
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, graph theory, and analytical geometry, will benefit from this discussion.

sarahr
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If the original graph, f, has a cusp, obviously the derivative is not defined at the x-value of the cusp (resulting in an asymptote).

but, what if you are viewing a graph of the derivative, f ', and it has a cusp.. what is going on at the x-value of the cusp on the original graph, f ?
 
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If the derivative graph has a cusp, that means that the second derivative does not exist. Think about y= |x| which does not have a derivative at x= 0. If we let f '(x)= |x| = (x if x>= 0 and -x if x< 0) and integrate we get
f(x)= ((1/2)x^2 if x>=0 and -(1/2)x^2 if x< 0). What does its graph look like around x= 0?
 
perfect example! thankss
 

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