Answer Key-hw1 (53)

ANSWER KEY ( HOMEWORK # 1 ) 1.) We must satisfy three conditions, namely : (i) 3 Kx R 0 which gives x % 3, (ii) 2K 3 Kx R 0 which gives x RK1, (iii) 1 K

2K 3 Kx R 0 which gives x % 2

Hence the domain is K1, 2

Domain : =,

2.)

3.) (a) We have f x = ln x C f Kx = ln

Kx C 1

=ln xC

2

Kx C1 =ln

xC

Range: All Integers R 0

x2 C1 =ln Kx C x2 C1

2

x C1

= ln

K1

=Kln x C

Kx C

x2 C1

2

x C1

xC

x2 C1

xC

x2 C1

=Kf x . Hence, f x is ODD.

2

x C1

x2 C1 , to solve for f K1 x , first interchange x and y and then e2 x K1 1 x K1 = e KeKx solve for y. After performing some algebra, we will get f x = x 2 2e (b) Since we have y = f x = ln x C

4.)

(a) By the Pythagorean Theorem, s = f d = d2 C36 ( d is in km. ) (b) If t is the time ( in hrs. ) that has elapsed since noon, d = g t = 30 t (c) f + g t = 900 t2 C36 This represents the distance between the ship and the lighthouse after t hours since noon.

5.) (a) FALSE - The logarithmic function loga x is a decreasing function for 0 ! a ! 1 (b) TRUE - We know h x = f g x . h Kx = f g Kx = f g x = h x . So h x is EVEN. (c) FALSE - f 6 or g 6 may not exist [ i.e. 6 is not in the domain of the functions ] but still lim f x $g x can exist.

x/6

x2 Cx K6 6.) To prove that lim = 5 we want to verify that c e O 0, d δ O 0 such that whenever x/2 x K2 x2 Cx K6 0 ! x K2 ! δ, then K5 ! e x K2 Upon considering a small δ Kneighborhood about 2 ( here x s 2 ), we have : x K2 x C3 x2 Cx K6 K5 = K5 = x C3 K5 = x K2 ! e x K2 x K2 This indicates that a possible choice for δ could be e. We show that this choice for δ works. Proof : Let e O 0. Let us choose δ = e Then 0 ! x K2 ! δ = e 0 x K2 ! e 0 x K2 = x C3 K5 =

x K2

x C3

x K2

x2 Cx K6 K5 = K5 ! e x K2

2

Hence, this shows that lim

x/2

x Cx K6 =5 x K2

7.) We want the circumference of our circle to be within 0.1 units from 6 π. So this means that our circumference is in the range of 6 π G0.1. We want to find the corresponding range of values for the radius 1 1 r. Now, 6 π G0.1 = 2 π r implies r = 3 G . So the radius must be within units from 3 so that 20 π 20 π 1 the circumference is within 0.1 units from 6 π ( This means that δ = ) 20 π