A useful remark: Since the derivative function is defined through limits, it is easy to show the following result: If f(s) and g(x) are equal on an open interval (a, b), and f?(c) exists for some c E (a, b), then g?(c) also exists and g(c) = f (c). this allows one to calculate the derivative of functions Defined piecewise, except at the x values where the definition changes from one formula to another. For example: if then h (x) = 4x on (- infinity , 5) (because it agrees with (2x^2 + 4)? there) and h?(x) = 2e^2x on (5, infinity) (because it agrees with (e^2x) there). Whether or not h?(5) exists requires more work and we will see how to answer this in the following two questions. (1) Consider the function (a) Show that f(s) is continuous at x = 0 by calculating the left and right hand limits of f(s) as it approaches 0 and showing they are both equal to f(0). (b) Show that f(s) is not differentiable at 0 by calculating and showing the two sided limit does not exist (note: calculate these limits explicitly, do not take shortcuts using derivative formulas). These one - sided limits are called the left and right - sided derivatives of f(x) at 0. By definition, f?(0) exists if and only if the two one - sided derivatives both exists and are equal (note: there is nothing special about 0; one sided derivatives are defined similarity for any s value). (c) Using the Power Rule Theorem from class and the remark preceding this question, calculate f(x) (note that by (b), 0 should not be in the domain of .f?(x)). Sketch a graph of f(s) and f(x) on the same XY - axis (you do not need to be precise, just sketch time shape).
