![]() While almost always the derivative is also differentiable, there is this counterexample: See 2013 AB 14 in which you must realize the since the function is given as differentiable at x = 1, it must be continuous there to solve the problem.Ī question that comes up is, if a function is differentiable is its derivative differentiable? The answer is no. Just remember: differentiability implies continuity. Just as important are questions in which the function is given as differentiable, but the student needs to know about continuity. On the AP Calculus exams students are often asked about the derivative of a function like those in the examples, and the lack of continuity should be an immediate clue that the derivative does not exist. This also applies to a situation like example 1 if f(3) were some value that did not fill in the hole in the graph. Therefore, even though the slopes from both side of x =1 approach the same value, namely 2, the derivative does not exist at x = 1. ![]() Since the point (1, 1) is on the left part of the graph, if h > 0, and the limit will always be a number greater than 3 divided by zero and will not exist. This function has a jump discontinuity at x = 1. ![]() So, in the limit definition of the derivative, there is no value of g(3) to use, and the derivative does not exist.Įxample 2. If a function is not continuous at a point, then it is not differentiable there.Įxample 1: A function such as has a (removable) discontinuity at x = 3, but no value there. This is a theorem whose contrapositive is used as much as the theorem itself. (The slope approaching from the left is not equal to the slope from the right.) A counterexample is the absolute value function which is continuous at the origin but not differentiable there. The converse of this theorem is false: A continuous function is not necessarily differentiable. Since both sides are finite, the function is continuous at x = a. The proof begins with the identity that for all If a function f is differentiable at x = a, then f is continuous at x = a. Function l(x) is continuous for all real values of x and therefore has no point of discontinuity.An important theorem concerning derivatives is this: ![]() Hence lim l(x) as x approaches -4 = 1 = l(-4). Function h is discontinuous at x = 1 and x = -1.ĭ) tan(x) is undefined for all values of x such that x = π/2 + k π, where k is any integer (k = 0, -1, 1, -2, 2.) and is therefore discontinuous for these same values of x.Į) The denominator of function j(x) is equal to 0 for x such that cos(x) - 1 = 0 or x = k (2 π), where k is any integer and therefore this function is undefined and therefore discontinuous for all these same values of x.į) Function k(x) is defined as the ratio of two continuous functions (with denominator x 2 + 5 never equal to 0), is defined for all real values of x and therefore has no point of discontinuity. The denominator is equal to 0 for x = 1 and x = -1 values for which the function is undefined and has no limits. Function g(x) is not continuous at x = 2.Ĭ) The denominator of function h(x) can be factored as follows: x 2 -1 = (x - 1)(x + 1). Therefore function f(x) is discontinuous at x = 0.ī) For x = 2 the denominator of function g(x) is equal to 0 and function g(x) not defined at x = 2 and it has no limit. A) For x = 0, the denominator of function f(x) is equal to 0 and f(x) is not defined and does not have a limit at x = 0.
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