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MATH 1300.03F D Exam 2 Monday, November, 8 1999

NAME:

Student Number:tex2html_wrap_inline153 No.tex2html_wrap_inline155Marks tex2html_wrap_inline157


tex2html_wrap_inline159 (4 Marks) Find the derivative of each of the following functions.

(1) tex2html_wrap_inline163.

Solution: tex2html_wrap_inline165.

(2) tex2html_wrap_inline169.

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)= 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.

tex2html_wrap_inline175

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

tex2html_wrap_inline177

This implies

tex2html_wrap_inline181

 

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 (-infty.bmp (654 bytes), -1)U(1,infty.bmp (654 bytes)).

(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)nequal.bmp (654 bytes)0 for xnequal.bmp (654 bytes)-1 and f''(-1) does not exist.

(3)We list the following table

x (- infty.bmp (654 bytes),-2) -2 (-2,-1) -1 (-1,0) 0

(0, infty.bmp (654 bytes))

f'(x) + 0 - DNE - 0 +
f''(x) -   - DNE +   +
f(x)  uparrow.bmp (654 bytes),down local max.  downarrow.bmp (654 bytes),down DNE  downarrow.bmp (654 bytes),up local min.  uparrow.bmp (654 bytes),up

f is increasing on (-infty.bmp (654 bytes), -2) and (0, infty.bmp (654 bytes)) and decreasing on (-2, -1) and (-1, 0). f is concave down on (-infty.bmp (654 bytes), -1) and concave up on (-1, infty.bmp (654 bytes)).

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

(5) a=limxarrow.bmp (654 bytes)infty.bmp (654 bytes) f(x)/x = x2/x2+x = 1 and b = limxarrow.bmp (654 bytes)infty.bmp (654 bytes)(f(x)-ax) = limxarrow.bmp (654 bytes)infty.bmp (654 bytes)( x2/1+x    -x)=-1. Hence, y=ax+b=x-1 is an asymptote. limxarrow.bmp (654 bytes)-1--f(x)=-infty.bmp (654 bytes) and

limxarrow.bmp (654 bytes)-1--f(x)=infty.bmp (654 bytes). 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.

kunquan.gif (3214 bytes)




Kunquan Lan
Mon Nov 8 10:45:34 EST 1999