# Calculus Questions with Answers (4)

 Calculus questions, on differentiable functions, with detailed solutions are presented. We first present two important theorems on differentiable functions that are used to discuss the solutions to the questions. Theorem(1): If function f is differentiable at x = a, then f is continuous at x = a. Contrapositive of the above theorem: If function f is not continuous at x = a, then f is not differentiable at x = a. Theorem(2): If function f is continuous at x = a and that lim f '(x) as x approaches a+ = lim f '(x) as x approaches a- then f is differentiable at x = a and f '(a) = lim f '(x) as x approaches a+ = lim f '(x) as x approaches a- Question 1: Determine whether function f defined by f(x) = 2 x 2      for x <= 1 and f(x) = 2 sqrt(x)    for x > 1 is differentiable at x = 1. Solution to Question 1: We shall use theorem 2 above to answer this question. We first need to investigate the continuity of f at x = 1. f(1)= 2 (1) 2 = 2 lim f(x) as x approaches 1- = 2 (1) 2 = 2 lim f(x) as x approaches 1+ = 2 sqrt(1) = 2 Since lim f(x) as x approaches 1- = lim f(x) as x approaches 1+ = f(1), function f is continuous at x = 1. Does f '(1) exists? For x < 1 we have f '(x) = 4 x For x > 1 we have f '(x) = 2 ( 1 / 2 ) ( 1/sqrt(x) ) = 1 / sqrt(x) We now calculate the limit of f '(x) as x approches 1 from the left and from the right. lim f '(x) as x approaches 1- = 4 (1) = 4 lim f '(x) as x approaches 1+ = 1 / sqrt(1) = 1 The two limits are not equal therefore f '(1) does not exit and f is not differentiable at x = 1. Question 2: Function f is defined by f(x) = x 3 for x <= 0 and f(x) = x 3 + 1 for x > 0. show that although lim f '(x) as x approaches 0+ = lim f '(x) as x approaches 0- f '(0) does not exist. Solution to Question 2: We first note that lim f (x) as x approaches 0- = 0 lim f (x) as x approaches 0+ = 1 The limits from the left and the right of 0 are not equal , therefore function f is not continuous at x = 0 and hence not differentiable at x = 0. We also note that f '(x) = 3x 2 for x <= 0 and f '(x) = 3 x 2 for x > 0. and lim f '(x) as x approaches 0+ = lim f '(x) as x approaches 0- = 0 In conclusion, it can be said that although lim f '(x) as x approaches 0+ = lim f '(x) as x approaches 0- f '(0) does not exist. Question 3: Find the values of A and B so that function f defined by f(x) = 2 x 2 for x <= 2 and f(x) = A x + B for x > 2 is differentiable at x = 2. Solution to Question 3: Let us first find the limits of f as x approaches 2 from the left and from the right. lim f (x) as x approaches 2- = 8 lim f (x) as x approaches 2+ = 2 A + b We now need to have f continuous at x = 2, hence 2 A + B = 8 We now calculate the derivative of f and find the limits as x approaches 2 from the left and from the right. f '(x) = 4x for x < 2 f '(x) = A for x > 2 and lim f '(x) as x approaches 2- = 8 lim f '(x) as x approaches 2+ = A For f to be differentiable at x = 2, we need to have lim f '(x) as x approaches 2- = lim f '(x) as x approaches 2+ Which gives A = 8 and using substitution in the equation 2 A + B = 8 to find B = - 8 Question 4: Find all values of x for which function f defined by f(x) = sqrt (x 2 - 2 x + 1) is not differentiable. Solution to Question 4: We first note that f(x) = sqrt (x 2 - 2 x + 1) = sqrt[ (x - 1) 2 ] = | x - 1 | f may be defined as follows f(x) = x - 1 for x > 1 and f(x) = - (x - 1) for x < 1 f is differentiable for all x < 1 and for all x > 1, we need to investigate its differentiability at x = 1. f is continuous at x = 1 because lim f (x) as x approaches 1- = lim f (x) as x approaches 1+ = 0 We now calculate the derivative of f and find its limits as x approaches 1 from the left and from the right. f '(x) = 1 for x < 1 and f '(x) = - 1 for x > 1 The limits of f ' as x approaches 1 from the left and from the right are not equal and therefore f is not differentiable at x = 1. In conclusion, f is differentiable everywhere except for x = 1. | 1 | 2 | 3 | 4 | 5 | More on calculus questions with answers, tutorials and problems .