MATH 1300.03F D Exam 2 Monday, November, 8 1999
Student Number: No.Marks
(4 Marks) Find the derivative of each of the following functions.
Solution: f'(x)= 4 cot3(5x2-2)[cot(5x2-2]' = 4 cot3(5x2-2)[ -csc2(5x2-2)(5x2-2)']
= -4 cot3(5x2-2)csc2(5x2-2).10x= -40x cot3(5x2-2)csc2(5x2-2).
2 (3 Marks) Find the second derivative of the following implicitly defined function.
Solution: Taking the derivative of this equation relative to x, we have 6x2+3y2y'=0
Hence, y'=-2x2/y2. Taking the derivative of (1), we obtain
3 (8 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)U(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=0f''(x)=2/(1+x)3. f''(x)0 for x-1 and f''(-1) does not exist.
(3)We list the following table
|f(x)||,down||local max.||,down||DNE||,up||local min.||,up|
f is increasing on (-, -2) and (0, ) and decreasing on (-2, -1) and (-1, 0). f is concave down on (-, -1) and concave up 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.
(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.