Determine the local extremum of the function?

Set the derivative equal to 0 and then see if the derivative is changing sign from before and after the x point where the derivative equals 0 (first derivative test). For example, lets say you have the function y=x^2. The derivative is dy/dx=2x. When does 2x=0? At x=0. This is called a critical point when dy/dx=0. To check if this point is a minimum or maximum check the derivative at a point just before the critical point. Check dy/dx at x=-1. This is -2. Check dy/dx at x=1. This is +2. So the derivative is changing therefore it is a point of local extrema. If the derivative is changing from positive to negative, there a local maximum at that point. If the derivative is changing from negative to positive, there is a local minimum at that point. This one is the latter (changing from negative to positive) so it is a local minimum. Try the same thing with the function that you have and it should work. However, this is a very basic example, so yours will probably be harder.

f(x) = e^x + e^(-x) f'(x) = e^x - e^(-x) f''(x) = e^x + e^(-x) to locate the extrema, enable f'(x) be 0 and remedy for x: 0 = e^x - e^(-x) e^(-x) = e^x ln(e^(-x)) = ln(e^x) -x*ln(e) = x*ln(e) -x = x x + x = 0 2x = 0 x = 0 to be certain what style of extremum happens at this element, plug x = 0 into the 2nd spinoff. If the 2nd spinoff is helpful, the element is a minimum. If detrimental, the element is a optimal. f''(0) = e^0 + e^(-0) = a million + a million = 2 So a minimum happens while x = 0. f(0) = e^0 + e^(-0) = a million + a million = 2 So the only extremum is a minimum at (0,2).