Hello

I'm having trouble figuring out the steps needed to solve the intersection between

f(x) = Cos (x)

f(x) = 1 - (2x/Pi)

I don't need the answer, only the steps used to solve such a problem. Thanks

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- February 19th 2011, 06:09 PMskyd171Finding intersection of trig and non trig function
Hello

I'm having trouble figuring out the steps needed to solve the intersection between

f(x) = Cos (x)

f(x) = 1 - (2x/Pi)

I don't need the answer, only the steps used to solve such a problem. Thanks - February 19th 2011, 06:25 PMmr fantastic
By inspection, x = 0 and x = pi are clearly solutions. There is one more solution but it cannot be found using algebra. In general equations involving such 'mixed functions' cannot be solved exactly using algebra, only approximate solutions using a CAS or numerical procedure can be found.

- February 19th 2011, 06:32 PMskyd171
Shame, I've spent about 4 hours trying to solve it with algebra only to find that out. This professor is a real piece of work.(Headbang)

- February 19th 2011, 06:38 PMmr fantastic
My previous post contains an error. The other solution is, again by inspection, x = pi/2.

If you are required to solve exactly equations like this without using technology, the solutions are generally designed to be so simple and obvious that they can be easily seen 'by inspection'. In fact, drawing accurate graphs would have put you on the right track and probably saved you 3 hours and 55 minutes of your time.

I have no problem with what your professor has asked you to do. - February 19th 2011, 06:48 PMskyd171
Yes, I used guess and check with graphs to find this solution. However I had been searching the web for hours for a proper algebraic way to find those roots and alternatively trying to find them numerically on my own. But it is just as informative to know that there is none.

This was only a small taste of my prof. The next problem (which I will not even bother to offend anyone here with) is

finding the area between 7 functions, the graphs of which look like a spider web. He is a swell guy.

thanks for the help - February 19th 2011, 06:53 PMmr fantastic