Manually Graphing Trigonometric Functions - Turning Points, Extrema

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SUMMARY

The discussion focuses on finding the turning points and extrema of the trigonometric function y1 = (20/x^2)sin(10/x) and its intersection with y2 = 5cos(x) within the interval (1,3). The user struggles with determining the turning points for y1, specifically where the derivative y'_1 equals zero. The solution involves setting up the equation and recognizing that numerical methods or graphing techniques may be necessary to find the turning points, as analytical solutions are complex.

PREREQUISITES
  • Understanding of calculus, specifically differentiation and finding derivatives.
  • Familiarity with trigonometric functions and their properties.
  • Knowledge of numerical methods for solving equations.
  • Ability to graph functions using graphing tools or software.
NEXT STEPS
  • Learn numerical methods for finding roots of equations, such as the Newton-Raphson method.
  • Explore graphing software like Desmos or GeoGebra to visualize trigonometric functions.
  • Study the properties of sine and cosine functions to better understand their behavior.
  • Investigate advanced calculus techniques for analyzing turning points and extrema.
USEFUL FOR

Students studying calculus, mathematicians interested in trigonometric functions, and educators seeking to enhance their teaching of graphing and analysis of functions.

phyzmatix
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Homework Statement



Two curves

[tex]y_1=(\frac{20}{x^2})\sin(\frac{10}{x})[/tex]

and

[tex]y_2=5\cos x[/tex]

intersect in three points in the interval (1,3). Draw the graphs, compute the minimum and maximum points as well as the turning points and show the points of intersection.

2. The attempt at a solution

The second function is straightforward, so my issues arise solely from y1. It's easy enough to see that y1 will be undefined at x=0 and also to determine the y-intercepts in the interval (1,3). My problems start when I try to determine the turning points for y1.

I know that the turning points will be found where

[tex]y'_1=0[/tex]

i.e. where

[tex]\frac{d}{dx}[(\frac{20}{x^2})\sin(\frac{10}{x})]=0[/tex]

[tex](\frac{-40}{x^3})\sin(\frac{10}{x})+(\frac{20}{x^2})\cos(\frac{10}{x})(\frac{-10}{x^2})=0[/tex]

[tex](\frac{-40}{x^3})\sin(\frac{10}{x})+(\frac{-200}{x^4})\cos(\frac{10}{x})=0[/tex]

[tex](\frac{-40}{x^3})[\sin(\frac{10}{x})+(\frac{5}{x})\cos(\frac{10}{x})]=0[/tex]

[tex]\sin(\frac{10}{x})=-(\frac{5}{x})\cos(\frac{10}{x})[/tex]

[tex]x\sin(\frac{10}{x})=-5\cos(\frac{10}{x})[/tex]

But now what? This can be simplified to either

[tex]\tan(\frac{10}{x})=\frac{-5}{x}[/tex]

or

[tex]x\tan(\frac{10}{x})=-5[/tex]

Neither of which help me much as I have no idea how to proceed beyond this point. Assuming I didn't make any mistakes, how do I determine the values of x where y1 will have turning points? Of course, if my reasoning is flawed/I made a mistake somewhere, please let me know.

Your help is appreciated!
phyz
 
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Your working looks OK. I think you may need to find the turning points using numerical methods or from the graph.
 
wywong said:
Your working looks OK. I think you may need to find the turning points using numerical methods or from the graph.

Well, I can't draw the graph until I have the turning points...But if you can enlighten me somewhat on how to go about determining the turning points using numerical methods I'll be extremely grateful!
 

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