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.

3. Show all of your work. Your work must justify the answer that you give.

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

(1) f(x) = 4x+(1-[1/x])2.

Solution: f(x) = [4/(2x)]+2(1-[1/x])[1/(x2)] = [2/(x)]+[(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)
cot(5x2-2)
= 4cot3(5x2-2)
-csc2(5x2-2) (5x2-2)
=
-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
- 2x2
y2

2

 

y2
=
-
4x+ 8x4
y3

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)(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) 0 for x -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.





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On 20 Apr 2001, 14:30.