Sketching the Curves of a Function W/In an Interval - Simple (1st Year Calcu

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SUMMARY

The discussion focuses on sketching the graph of the function y = cos(x) - 1/2(cos(2x)) over the interval [0, 2π]. The user initially identifies the absolute minimum at (π, -3/2) but struggles to locate other relative extrema. The first derivative, -sin(x) + sin(2x), is used to find potential extrema at 0, π, and 2π. The user resolves their confusion by realizing the need to apply the double angle trigonometric formula, correcting their earlier algebraic oversight.

PREREQUISITES
  • Understanding of trigonometric functions and their properties
  • Knowledge of calculus concepts such as derivatives and extrema
  • Familiarity with graphing functions over specified intervals
  • Ability to apply double angle formulas in trigonometry
NEXT STEPS
  • Study the application of the double angle formulas in trigonometric identities
  • Learn how to find relative and absolute extrema using calculus
  • Practice sketching graphs of trigonometric functions over various intervals
  • Explore the concepts of concavity and points of inflection in function analysis
USEFUL FOR

Students in first-year calculus, particularly those learning to analyze and sketch trigonometric functions, as well as educators seeking to clarify concepts of extrema and graphing techniques.

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[SOLVED] Sketching the Curves of a Function W/In an Interval - Simple (1st Year Calcu

Homework Statement



Sketch the graph of the function on the interval [0, 2pi].

y = cosx - 1/2(cos2x)

Homework Equations



0320.png


The Attempt at a Solution



so the problems that i have been practicing like this have been pretty simple. i determine all of the following:

  1. domain and range
  2. x and y intercepts
  3. whether or not there is discontinuity
  4. whether or not there is symmetry
  5. intervals of increasing and decreasing order
  6. extrema
  7. points of inflection
  8. concavity
  9. whether or not there exist asymptotes

the problem is that mathematically i located a single extreme value at (pi, -3/2), but graphically, there appears to be more.

http://img149.imageshack.us/img149/1645/graphscreenshotth2.th.png

is not each change in direction a relative minimum/maximum value? i can only seem to locate the absolute minimum value within the interval, but none of the other relative extreme values. any idea as to where i am going wrong?

so, the first derivative of the given function is -sinx + sin2x. solving this equation is where i find possible extrema at 0, pi, and 2pi. any ideas why i can't seem to locate the other extrema algebraically?
 
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so, the first derivative of the given function is -sinx + sin2x. solving this equation is where i find possible extrema at 0, pi, and 2pi.

How did you solve this equation?
 
ah. figured it out. i had worked out the values mentally real quick and hadn't realized that i needed to use the double angle trig formula. sheesh. yet another simple algebra mistake. is there a way i could mark this for deletion or something? by the way, thank you. sometimes it can be difficult to see a mistake of your own you know.
 
You are welcome! :smile:

Just edit the title and marked it [SOLVED]
 

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