# Manually Graphing Trigonometric Functions - Turning Points, Extrema

#### phyzmatix

1. The problem statement, all variables and given/known data

Two curves

$$y_1=(\frac{20}{x^2})\sin(\frac{10}{x})$$

and

$$y_2=5\cos x$$

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

$$y'_1=0$$

i.e. where

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

$$(\frac{-40}{x^3})\sin(\frac{10}{x})+(\frac{20}{x^2})\cos(\frac{10}{x})(\frac{-10}{x^2})=0$$

$$(\frac{-40}{x^3})\sin(\frac{10}{x})+(\frac{-200}{x^4})\cos(\frac{10}{x})=0$$

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

$$\sin(\frac{10}{x})=-(\frac{5}{x})\cos(\frac{10}{x})$$

$$x\sin(\frac{10}{x})=-5\cos(\frac{10}{x})$$

But now what? This can be simplified to either

$$\tan(\frac{10}{x})=\frac{-5}{x}$$

or

$$x\tan(\frac{10}{x})=-5$$

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

#### wywong

Your working looks OK. I think you may need to find the turning points using numerical methods or from the graph.

#### phyzmatix

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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