Final Review

Differential Calculus. Final review. I. Find the given limits.

t 3 − 27 . t2 −9 sin 2 x (b) lim x →0 . sin 5 x 4 − x2 lim (c) . x →2 3 − x2 + 5 (d) lim x → a 3 g ( x) [ f ( x) + 3] , if lim x →a f ( x ) = 3 and lim x→ a g ( x ) = −1 . (a)

lim t →3

II. Find the vertical and horizontal asymptotes of

y=

x2 + 4 . x2 −1

III. Find the derivative of the following functions (a)

(

−x

e +e . e x − e−x    1 (c) y = cos sin  tan  x   4 1+ ex  . (d) f ( x) = ln x  1− e  (b)

(e)

)

f (θ ) = sin (θ ) 1 − cos 2 (θ ) ( cot (θ ) sec(θ ) ) . y=

x

      . 

e xy + x 2 − y 2 = 10 . (Implicit differentiation).

IV. Sketch the graphs of the following functions, using the first and second derivative rules.

17. f ( x) = 1 − 3 x + 3 x 2 − x 3 . 19. f ( x) = x 4 − 6 x 2 . 21. f ( x) = ( x − 3) . 4

V. In the production and selling of x units of a certain merchandise, the function price p and the function cost C (in dollars) are given by

p ( x ) = 5.00 − 0.002 x C ( x) = 3.00 + 1.10 x

Find the expressions for the marginal income, cost and revenue; find the level of production that gives the maximum revenue. VI. An open rectangular box is to be constructed by cutting square corners out of a 16- by 16inch piece of cardboard and folding up the flaps. [See Fig. 10(b).] Find the value of x for which the volume of the box will be as large as possible.

VII. A toy rocket fired straight up into the air has height s (t ) = 160t − 16t feet after t seconds. (a) What is the rocket’s initial velocity (when t = 0)? (b) What is the velocity after 2 seconds? (c) What is the acceleration when t = 3? (d) At what time will the rocket hit the ground? (e) At what velocity will the rocket be traveling just as it smashes into the ground? 2

Answers: I. (a) 9/2, (b)

2 /5, (c) 6, (d) -6.

II. Vertical Asymptotes: x = 1 and x = -1, Horizontal Asymptote: y = 1.

III. (a)

(c)

f (θ ) = 2 sin(θ ) cos(θ ) , (b) y ' =

   1 y ' = sin  sin  tan x   4

− ye xy − 2 x (e) y ' = . xe xy − 2 y V.

(e

     1    cos tan x    4

−4 x

− e −x

)

2

,

 2  1   sec  x  4

 ln 4 2e x , (d) ,  x f ' ( x) =  4 1 − e2x

IV. Income:

xp( x) = 5.00 x − 0.002 x 2 , marginal income: 5.00 − 0.004 x ,

Marginal cost:

C ' ( x) = 1.10 .

Revenue: Income – Cost:

5.00 x − 0.002 x 2 − ( 3.00 + 1.10 x ) = −3.00 + 3.9 x − 0.002 x 2 .

Marginal revenue: 3.9 − 0.004 x . Maximum revenue:

x = 975 .

VI.

V = w2 x w = 16 − 2 x 2 V = (16 − 2 x ) x

Objective equation.

Volume:

(

)

V = 256 − 64 x + 4 x 2 x V = 256 x − 64 x 2 + 4 x 3 V ' = 256 − 128 x + 12 x 2 V '= 0

Constraint equation. We substitute the expression for w (constraint equation) in the expression for the volume (objective equation). That way, the expression for the volume depends on just one variable, x. We want to find the maximum of the function for the volume, so we find its derivative, and set it equal to zero.

256 − 128 x + 12 x 2 = 0 64 − 32 x + 3 x 2 = 0 x=

− ( − 32 ) ±

( 32) 2 − 4( 3)( 64) 2( 3)

32 ± 16 x= 6 x1 = 8, x 2 ≈ 2.7.

We can’t solve this equation using factorization, so we use the general formula for second degree equations:

x=

− b ± b 2 − 4ac . 2a

Finally, we choose x = 2.7 as our final answer, the other value (x = 8), is a minimum, because if we evaluate the second derivative of the volume ( V ' ' = −128 + 24 x ) with x = 8, we obtain a positive value, which means that the function is concave up at x = 8.

VII. (a) 160 ft/sec (b) 96 ft/sec (c) -32 ft/sec2 (d) 10 sec (e) -160 ft/sec.