Fixed Point Theorem
Okay I have a homework question that is kinda tricky (to me anyway). I am supposed to show that the nonlinear equation f(x) = e^-x - cosx = 0 has a solution on the interval I = [0.5,4]. The method we were taught was the Fixed Point theorem. It says
Let I = [a, b]. Suppose g(I) ⊆ I and g(x) is continuous on I. Then g(x) has at least a fixed
point in I. The function, when plotted on a calculator, was continuous on the interval.
The work so far...
In class we found |derivative of f_1| and saw if it was less than 1.
f(x) ==> f_1(x) = -ln(cosx)....then the derivative of f_1 = sinx/cosx. So What I did was that I plugged in .5 and 4 into the derivative separately and it showed that it was less than one.
I am not entirely sure as to what is happening here. If you can just explain the Fixed Point theorem I would be grateful and how I could use it to solve this problem. It's just that in class we found the absolute value of the derivative and looked to see if it was less than 1. Not sure of the point there. Thanks.
Originally Posted by mesmo