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

NAME:

Student Number: No.Marks

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

(1) .

Solution: .

(2) .

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

(3) f(x)=sec^{2}(x+1)

Solution: f'(x)=2sec(x+1)(sec(x+1))'=2sec(x+1)tan(x+1)sec(x+1)=2sec^{2}(x+1)tan(x+1).

(4) f(x)=cot^{4}(5x^{2}-2)**.**

Solution: f'(x)= 4 cot^{3}(5x^{2}-2)[cot(5x^{2}-2]' = 4 cot3(5x^{2}-2)[
-csc2(5x^{2}-2)(5x^{2}-2)']

= -4 cot^{3}(5x^{2}-2)csc^{2}(5x^{2}-2).10x= -40x cot^{3}(5x^{2}-2)csc^{2}(5x^{2}-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 6x^{2}+3y^{2}y'=0

Hence, y'=-2x^{2}/y^{2}. *Taking the derivative of* (1), *we obtain*

This implies

**3** (8 *Marks*) *Let* f(x)=x^{2}/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

x | (- ,-2) | -2 | (-2,-1) | -1 | (-1,0) | 0 | (0, ) |

f'(x) | + | 0 | - | DNE | - | 0 | + |

f''(x) | - | - | DNE | + | + | ||

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=lim_{x} f(x)/x = x^{2}/x^{2}+x = 1 *and* b
= lim_{x}(f(x)-ax) = lim_{x}( x^{2}/1+x
-x)=-1. *Hence*, y=ax+b=x-1 is an asymptote. lim_{x-1}^{-}-f(x)=- *and*

lim_{x-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.

Mon Nov 8 10:45:34 EST 1999