Inverse Functions: Arcsec(x)
Date: 02/11/99 at 20:15:09 From: Knote Subject: Calculus If y = arcsec(x) then what does x equal?
Date: 02/22/99 at 15:53:11 From: Doctor Elizabeth Subject: Re: Calculus Hi, Knote. Good question! Let's take a closer look at that. First, what does the notation "arcsec()" stand for? Well, obviously it has something to do with trigonometry. More specifically, judging from the second half - "sec" - you might already know that it is somehow related to the sec() value of an angle. Now the question is, what is the relation? Let me give you a hint: arcsec(x) is the INVERSE function of sec(x). Have you studied inverse functions yet? If not, I'll give you a quick overview. Basically, if y = f(x), then f^(-1)(x) is how we represent its inverse. Remember the definition of a function? It's like a mapping. Each value of x has a corresponding value y, after some operation which is represented by f. Its inverse, then, as you would expect, expresses the relation the other way round. For each value of y, we look for an x, and the operation we need to go through to get from y->x is the inverse of f, or f^(-1). For example, we have the function y = 3x. When x=1, y=3. Now when y=3, what's value of x? x=1, right? And so how is x related to y? x=(1/3)y, which is the inverse function we are looking for! By convention, we use x as the independent variable, so we write f^(-1)(x)=(1/3)x. So what's special about inverse functions? If we didn't change the independent variable in the inverse function, what would it be? Isn't it x = f^(-1)(y)? And the original function is y=f(x). Now have you noticed something in common in these two expressions? How about a substitution? Let's write y as f(x) in the inverse function. Now we get x = f^(-1)(f(x))*. Back to your question: arcsec(x) is the inverse of sec(x). If we plug them in into the equation we just got(*), we'll have x = arcsec(sec(x)) or x = sec(arcsec(x)), depending on whether you use sec(x) as the original function or the other way round. I'll leave this part for you, then. Good luck in figuring out this last step. If you still have questions, don't hesitate to write us back. - Doctor Elizabeth, The Math Forum http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum