Continuity and Differentiability of Piecewise FunctionDate: 05/04/98 at 11:25:23 From: Heather Hopson Subject: differentiable and continous functions A function is defined by the following formula: f(x) = x^2 + 2, for x less than or equal to 1 = a(x - (1/x)) + b, for x greater than 1 Find a and b such that f is continuous and differentiable. Plot the function, if possible. How do I find a and b? If it is continuous, isn't it differentiable? Date: 05/04/98 at 15:53:39 From: Doctor Sam Subject: Re: differentiable and continous functions Heather, Let's start with your last question first. The answer is no. A function can be continuous without having a derivative. The absolute value function is a good example of this: y = |x| (the graph looks like a V). It is continuous but it has no slope at the origin, at the corner of the V. Differentiable means it must have a derivative, a slope, at every point. When a graph has a derivative at a point, it is sometimes called "locally linear." That's because if you zoom in very close to the point, the graph appears to be a straight line. Similarly, the Earth looks flat to us, living on its surface. But from space it appears curved. A graph like y = x^2 will appear to be a straight line if you look at it under a microscope. Thus, it is differentiable. The absolute value graph, however, will always look like a V, no matter how much you zoom in towards the vertex (0,0). You will never straighten the graph out. So at x = 0, y = |x| is NOT locally linear and, therefore, does not have a derivative. The reverse proposition, by the way, is true. That is, if a function is differentiable it MUST be continuous. If, when you zoom in on the graph it appears to be a straight line, then it is continuous, since straight lines are continuous. Now on to the rest of your question: f(x) = x^2 + 2, for x less than or equal to 1 = a(x - (1/x)) + b, for x greater than 1 f(x) is an example of a "piecewise function." That is, it is made up of two pieces. One piece is a part of the parabola y = x^2 + 2. The other piece is part of a weird curve, a(x - (1/x)) + b. Depending upon the choice of "a" and "b," the two pieces need not join together, in which case the curve will not be continuous and, therefore, not differentiable. On the other hand, there may be choices of "a" and "b" that will let the pieces join together -- smoothly, or, like the absolute value function, the pieces may form a corner, in which case it will not be differentiable, since the function won't be locally linear at the point of joining. That is what you are seeking: smooth curves are differentiable. So let's take the first part first. What values of "a" and "b" will make this function continuous? We must find values that will let the two pieces of the graph connect. Recall the function: f(x) = x^2 + 2, for x less than or equal to 1 = a(x - (1/x)) + b, for x greater than 1 The key point here is the value x = 1. That is where f(x) breaks in two pieces. The first part of the definition shows that f(1) = 3, so the first piece of the graph ends at the point (1, 3). If the function is continuous, then the second piece of the graph must start at (1,3). That is, when x = 1, we need a(1 - (1/1)) + b = 3 Thus, b = 3. So if b = 3, the two pieces will connect. That is true for ANY value of "a." If you have a graphing calculator or graphing software, you might try graphing y = x^2 + 2 and y = a(x - (1/x)) + 3 for several different values of "a." In every case, the graphs intersect at (1,3), but at different angles. Now we are ready for the last part of the question. What value of "a" will make the function differentiable? That means we want to avoid a corner at (1,3). To do this, we need to find the derivative of each part of the function: When x < 1, f'(x) = 2x. At x = 1, this piece has slope 2. When x > 1, f'(x) = a(1 + 1/x^2). At x = 1, this piece has slope a(1 + 1) = 2a. You see, different choices of "a" will give different slopes for this piece. To make f(x) differentiable at (1,3), we want the two pieces to have the same slope. This gives us an equation: 2 = 2a So choosing a = 1 and b = 3 will produce a function that is both continuous and differentiable. Choosing any other value for "a" with b = 3 will give a function that is continuous but not differentiable at x = 1. Finally, any other choice of "a" and "b" will produce a function which is discontinuous at x = 1 and, therefore, not differentiable there either. I hope that helps. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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