Derivatives and Continuity

Transcription

1 Derivatives and Continuity As with limits, derivatives do not exist unless the right and left-hand derivatives both exist and are equal. There are three main instances when this happens: One, if the curve is not continuous at a point, the derivative does not exist because there will be no slope on one side of the point. Two, if the curve makes an abrupt turn (a cusp or a corner) the slopes on either side of the point will not be equal. or

2 Three, if there is a vertical tangent at a point the derivative is undefined (does not exist) because the derivative gives the slope of the tangent and that slope will be undefined for a vertical line. Go over them carefully; you will be expected to be able to look at a graph and determine where the derivative exists. These illustrations also emphasize one very important point. Differentiability implies continuity (the function must be continuous in order for a derivative to exist), but continuity does not imply differentiability (as in examples and 3 a continuous function still might not have a derivative if there is a cusp or vertical tangent). Just because the curve is continuous, it does not mean that a derivative must exist. This is a perfect example, by the way, of an AP exam question!

3 Local Linearity The term local linearity refers to the fact that a smooth curve that is differentiable will closely resemble its tangent line IF you zoom in close enough. We will use this later in the course with linear approximations, where we find the tangent line and use it instead of the function to estimate y-values on the curve close to the point of tangency. It is important to be able to visualize this, so do this graphing exercise now. ) Graph the function y = x + x in a normal window. ) Zoom in on x =.5 until the curve looks like a straight line. When you get to the point where the curve looks straight, you essentially have the graph of the tangent line as well. Close to the point of tangency the function and its tangent are the same, and that fact is what lays the groundwork for linear approximations.

4 Evaluating Derivatives with the Calculator One of the things you will be expected to do with your calculator in this course is to evaluate a derivative at a particular point using your calculator. We of course can do this by first finding the derivative the long way and then evaluating it at our x-value, but if all we need is the final value we can do it in one step with the calculator! Don't get too excited though you will get to do this on some long answer free-response questions but for most of time you will be required to show the derivation process. Here is how to evaluate a derivative without first finding the derivative. 3 Suppose I want to find the velocity at t = 6, for the position function st ( ) 4t 5t 45t 30. Instead of first finding the derivative, s '(t) = v (t), and then evaluating that derivative at t = 6, we can do it in one step on the calculator. () Go to the MATH menu and choose nderiv( function. On the TI-8 and 83+ it is #8 in the math menu. () Type in function, variable, value you want to evaluate at. For our example you would type in this: 4t^3+5t^-45t+0, t, 6 and then press enter. The answer will be 46,703. Make sure you can do this, so try it now on your calculator. You will most frequently have to find the derivative in this class, since that is one of the things we need to practice, but there will be times when you can do it the short way with the calculator. The trick is to make sure to use proper notation, because without it you will not get credit for your answer. For the problem we just did, you would need to write s '(6) = v (6) = 46,703. That tells the reader that if you find the derivative of the position function, you will get the velocity function, and that when you evaluate that at t = 6, your answer is 46,703 ft/s (or whatever the units are).

5 Homework Examples #5, 7, 9 #5: For the given graph of a function over the closed interval 3 x, at what domain points does the function appear to be: a) differentiable? b) continuous but not differentiable? c) neither continuous nor differentiable? -3 a) Differentiable for all points on (-3, ) The derivative exists along the entire curve except at the endpoints. At the endpoints there is only a onesided derivative. b) Only at the endpoints, x = -3,, is the The function is everywhere continuous because it is curve continuous but not differentiable. c) There are no points where the curve is neither continuous nor differentiable. everywhere differentiable. The function is everywhere continuous and differentiable.

6 #7: For the given graph of a function over the closed interval 3 x 3, at what domain points does the function appear to be: a) differentiable? b) continuous but not differentiable? c) neither continuous nor differentiable? -3 3 a) Differentiable for all points on (-3, ) except for x = 0. b) The curve is continuous but not differentiable at the endpoints, x = -3, 3. c) At x = 0, the curve is neither continuous nor differentiable. The derivative exists along the entire curve except at the point of discontinuity and the endpoints. There is no point where the function is not differentiable but still continuous. At the break in the curve, the function is neither continuous nor differentiable #9: For the given graph of a function over the closed interval x, at what domain points does the function appear to be: a) differentiable? b) continuous but not differentiable? c) neither continuous nor differentiable? - a) Differentiable for all points on (-, )] except for x = 0 and the endpoints. b) At x = 0, -, the curve is continuous but not differentiable. c) There is no point where the curve is neither continuous nor differentiable. The derivative exists along the entire curve except at the cusp and the endpoints. The function is continuous at the cusp and endpoints even though it is not differentiable. There are no discontinuities on the curve.

7 Homework Examples #, 3, 7 tan x, x 0 #: The function y is not differentiable at x = 0. Tell whether the problem, x 0 is a corner, cusp, vertical tangent, or discontinuity. tan x, x 0 y, x 0 tan 0 0 and f (0) = There is a discontinuity. This is our function. First, check to see if the function is continuous (it is the easiest thing to check). Both pieces are continuous, so the only possible discontinuity would be where the pieces meet. #3: The function y x x corner, cusp, vertical tangent, or discontinuity. is not differentiable at x = 0. Tell whether the problem is a y x x y x x, x 0 y x x 0 There is no discontinuity because to the left of x = 0, y = and at x = 0, y =. 0, x 0 y x 0 There is a corner. This is our function. We can rewrite it since we are looking for the positive square root. Now, we are rewrite it again using a piecewise definition. The x = x when x is positive and x = -x when x is negative. First, check to see if the function is continuous (it is the easiest thing to check). Both pieces are continuous, so the only possible discontinuity would be where the pieces meet. Now look at the slopes on either side of x = 0 to see if they are equal. If they are not, the curve is not smooth. A corner is a sharp turn. A cusp will have infinite slopes on either side, but going in different directions. ** you can also see this on the graph, and for now that is an ok way to answer these questions.

8 #7: Find all values of x for which the function f( x) 3 x 8 x 4x 5 is differentiable. 3 x 8 f( x) x 4x 5 x - 4x - 5 = (x - 5)(x + ) The denominator equals zero when x = 5, - and thus the function will be undefined at those points and there will be discontinuities. The function is differentiable for all x 5, This is our function. Both the numerator and denominator consist of polynomials that are everywhere continuous. Thus, the only points where we might not have a derivative would be values of x where the function is not defined. So, look for values that would make the denominator equal to zero. Those values must be excluded from the domain, and a function is not differentiable at an discontinuity.