Quarter CirclesDate: 12/20/96 at 22:07:38 From: Jason k Cheung Subject: Functions How would I make a function that would graph a quarter of a circle in the quadrants II and IV? Date: 12/21/96 at 03:33:24 From: Doctor Pete Subject: Re: Functions One possible way to do this is as follows: We know we can draw a semicircle in quadrants I and II with: f(x) = +Sqrt[1-x^2] (i.e., the postive square root of 1 minus x squared). This gives the correct graph in quadrant II, but we would like to "flip" the graph about the x-axis for x > 0. That is, if g(x) is our desired function, then g(x) = -f(x) for x > 0, and g(x) = f(x) for x < 0. To do this, consider the function h(x) = |x|/x, that is, the absolute value of x divided by x. For x > 0, this is 1, and for x < 0, this is -1. (Question: What is h(0)? This is of interest, because your answer to this will affect what g(0) will be.) Now, I leave it to you to figure out how to apply h(x) to f(x) to get g(x). -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/24/96 at 15:10:41 From: Jason k Cheung Subject: Re: Functions Is this what you were thinking about?: f(x)=sqrt[1-(sqrt(-x))^4] This would give me a graph in quadrant II. f(x)=-sqrt[1-(sqrt(x))^4] This would graph quadrant IV. Date: 12/28/96 at 20:09:59 From: Doctor Pete Subject: Re: Functions Hmm.... while I might agree that the equations you have given are technically correct, I think I should point out a few things. First, note that Sqrt[-x]^2 = Sqrt[x]^2 = |x|, the absolute value of x. Then Sqrt[-x]^4 = Sqrt[x]^4 = |x|^2 = x^2, because the square of a number is always non-negative. So there is no sense in differentiating between the two; that is, for all values of x: Sqrt[1-Sqrt[-x]^4] = Sqrt[1-Sqrt[x]^4] = Sqrt[1-x^2] Second, the purpose of introducing the function h(x) = |x|/x is to try to do what you have done by splitting up g(x) into two cases. (I will call g(x) the function that we wish to obtain, and f(x) = Sqrt[1-x^2] as I have defined in the previous message.) In order to get the desired function, we could have simply written: / f(x) = Sqrt[1-x^2] , x < 0 g(x) = -| \ -f(x) = -Sqrt[1-x^2] , x > 0 This definition is not always desirable. You can use h(x) to do what the minus sign is doing in the second case, because h(x) = 1 when x > 0, and -1 when x < 0. So -h(x) = -1 when x > 0, and -h(x) = 1 when x < 0. This gives you the correct sign, so in order to merge the two cases into one, we take the product -f(x)h(x), which gives: g(x) = -f(x)h(x) = -Sqrt[1-x^2] |x|/x This is the function we're looking for. If x < 0, then |x|/x = -1, and g(x) = Sqrt[1-x^2]. If x > 0, then |x|/x = 1, and g(x) = -Sqrt[1-x^2]. Also note that g(x), the function we were looking for, is discontinuous at 0, and is defined on the domain [-1,0) U (0,1]. f(x) = Sqrt[1-x^2], however, is defined, and continuous, on the entire interval [-1,1]. So obtaining g(x) from f(x) requires the introduction of a function h(x) that has a discontinuity at 0. Indeed, h(x) = |x|/x is not continuous at 0, however you may wish to define h(0). -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/