Exam 2 Exam 2                              Friday, April 20, 2001

NAME:

Student Number:------            No.----           Marks ----

Instructions:

1. You have 50 minutes for this exam. You are not allowed to use any calculators.

2. This exam contains 3 questions and has a total of 100 marks.

1 (30 Marks) Find the derivative of each of the following functions.

(1) f(x) = 4Öx+(1-[1/x])2.

Solution: f¢(x) = [4/(2Öx)]+2(1-[1/x])[1/(x2)] = [2/(Öx)]+[(2(x-1))/(x3)].

(2) f(x) = cos(5x+3).

Solution: f¢(x) = -sin(5x+3)(5x+3)¢ = -5sin(5x+3).

(3) f(x) = sec2(x+1).

Solution: f¢(x) = 2sec(x+1)(sec(x+1))¢ = 2sec(x+1)tan(x+1)sec(x+1) = 2sec2(x+1)tan(x+1).

(4) f(x) = cot4(5x2-2).

Solution:

 f¢(x)
 =
 4cot3(5x2-2) éë cot(5x2-2) ùû ¢ = 4cot3(5x2-2) éë -csc2(5x2-2) (5x2-2)¢ ùû
 =
 -4cot3(5x2-2)csc2(5x2-2).10 x = -40xcot3(5x2-2)csc2(5x2-2).

2 (20 Marks) Find the second derivative of the following implicitly defined function.

 2x3+y3 = 1.

Solution: Taking the derivative of this equation relative to x, we have 6x2+3y2y¢ = 0, that is,

 2x2+y2y¢ = 0
(1)

Hence, y¢ = -[(2x2)/(y2)]. Taking the derivative of (1), we obtain

 4x+2y(y¢)2+y2y¢¢ = 0.

This implies

 y¢¢
 =
- 4x+2y(y¢)2
y2
= -
 4x+2y æè - 2x2y2 öø 2

y2
 =
-
 4x+ 8x4y3

y2
= - 4xy3+8x4
y5
.

3 (50 Marks) Let f(x) = [(x2)/(1+x)].

(1) Find the domain of f.

(2) Find all the critical points of f and the second derivative of f.

(3) Find the intervals on which f is increasing, decreasing, concave down or concave up.

(4) Evaluate the local maximum and local minimum values of f.

(5) Find all the asymptotes of f.

(6) Find the x-intercepts and y-intercept of f.

(7) Sketch the graph of f.

Solution: (1) The domain of f is (-¥, -1)È(1,¥).

(2) f¢(x) = [(x(x+2))/((1+x)2)]. Let f¢(x) = 0. Then x = 0 and x = -2. Moreover, f¢(-1) does not exist. Hence, all the critical points are x = -1, x = -2 and x = 0.

f¢¢(x) = [2/((1+x)3)]. f¢¢(x) ¹ 0 for x ¹ -1 and f¢¢(-1) does not exist.

(3) We list the following table.

 x
 (-¥, -2)
 -2
 (-2, -1)
 -1
 (-1,0)
 0
 (0, ¥)
 f¢(x)
 +
 0
 -
 DNE
 -
 0
 +
 f¢¢(x)
 -
 -
 -
 DNE
 +
 +
 +
 f(x)
 \nearrow, up
 local max.
 \searrow, up
 DNE
 \searrow, down
 local min.
 \nearrow, down

f is increasing on (-¥, -2) and (0, ¥) and decreasing on (-2, -1) and (-1, 0). f is concave up on (-¥, -1) and concave down on (-1, ¥).

(4) The local maximum value f(-2) = -4 and the local minimum value f(0) = 0

(5) a = limx®¥[f(x)/x] = [(x2)/(x2+x)] = 1 and b = limx® ¥(f(x)-ax) = limx® ¥([(x2)/(1+x)]-x) = -1. Hence, y = ax+b = x-1 is an asymptote.

limx® -1-f(x) = -¥ and limx® -1-f(x) = ¥. Hence, x = -1 is a vertical asymptote.

(6) x = 0 is the x-intercept of f and f(0) = 0 is the y-intercept of f.

(7) Sketch the graph of f as follows.

File translated from TEX by TTH, version 2.79.
On 20 Apr 2001, 14:30.