Math 1851 (a)

MATH 1851 (A)

Calculus and Ordinary Differential Equations

Test 1

Date: 29th September, 2014 Time allowed: 45 minutes Name: University Number:

Answer ALL THREE questions 1. (40%) Let y = f (x) = 2x3 − 9x2 + 12x + 1. (a) Sketch the graph of the function y = f (x) for 0 ≤ x ≤ 3. (b) Find the absolute maximum and absolute minimum attained by y for 0 ≤ x ≤ 3. (c) Is the function y = f (x), 0 ≤ x ≤ 3 one-to-one? Why? Solution. (a) f ′ (x) = 6x2 − 18x + 12 = 6(x − 2)(x − 1). Therefore the turning points of y = f (x) are at x = 1, 2. It is easy to see that f ′ (x) < 0 for 1 < x < 2 and f ′ (x) > 0 elsewhere. Thus, y = f (x) attains local maximum at x = 1 with f (1) = 6 and f (x) attains local minimum at x = 2 with f (2) = 5. At the end points x = 0, 3, we have f (0) = 1 and f (30) = 10. (b) From the analysis and the sketch in part (a), we find that in the range 0 ≤ x ≤ 3, f (x) attains absolute maximum value of 10 at x = 3 and absolute minimum value of 1 at x = 0. (c) The function y = f (x) for 0 ≤ x ≤ 3 is not one-to-one. To show this, one just has to find a value c such that at least two distinct values x1 , x2 in the interval [0, 3] satisfy f (x1 ) = f (x2 ) = c. For example, the value c = 5 works. First x1 = 2 satisfies f (x1 ) = 5. Moreover, since the function y = f (x) is a continuous function (it is a polynomial) and f (0) = 1, f (1) = 6, by the intermediate value theorem, there must be at least one value of x2 in the interval [0, 1] such that f (x2 ) = 5. Clearly x1 ̸= x2 and this proves that f (x) is not one-to-one. 2. (30%) Find the following limits. (You need to state clearly the theorems you use) (a) limx→0 (1 + x) x Solution. Let y = (1 + x) x . Then ln y = lim ln y = lim

1 1

1 x

ln(1 + x). We have,

ln(1 + x) (1 + x)−1 = lim = 1. x→0 x→0 x→0 x 1 Here in the last step, we have applied the l’Hˆpital rule. Thus, limx→0 y = elimx→0 ln y = o 1 e = e. h2 (b) limh→0 1−cos h Solution. We apply the l’Hˆpital rule twice (we have the indeterminate form 0 )...