Calculus 221 - First Exam (50 Minutes).pdf

Calculus 221 - First Exam (50 Minutes)
Friday October 4 1996

I Find the limit or show that it does not exist. Justify your
answer.
t3 − 1
(1) lim 2
t→−∞ t − 1
Answer:
1
t3 − 1 t− 2
lim = lim t = −∞
t→−∞ t2 − 1 t→−∞ 1
1− 2
t

t3 − 1
(2) lim
t→3 t2 − 1
Answer:
t3 − 1 33 − 1 26
lim = 2 =
t→3 t −1
2 3 −1 8

t3 − 1
(3) lim
t→1 t2 − 1
Answer:
t3 − 1 (t − 1)(t2 + t + 1) (t2 + t + 1) 3
lim = lim = lim =
t→1 t −1
2 t→1 (t − 1)(t + 1) t→1 (t + 1) 2



II Find the limit or show that it does not exist. Justify your
answer.
sin(3x)
(1) lim
x→∞ x
Answer: Since −1 ≤ sin(3x) ≤ 1 for all x we have
1 sin(3x) 1
− ≤ ≤
x x x
sin(3x)
for x > 0. Hence lim = 0 by the Squeeze theorem.
x→∞ x
sin(x2 )
(2) lim
x→0 x
Answer:
sin(x2 ) sin(x2 ) sin(x2 ) sin(u)
lim = lim 2
· x = lim · lim x = lim · lim x = 1 · 0 = 0.
x→0 x x→0 x x→0 x2 x→0 u→0 u x→0



sin(x) − sin(π/3))
(3) lim
x→π/3 x − π/3

,Answer: Let f (x) = sin(x) and a = π/3. Then f 0 (x) = cos(x) and

sin(x) − sin(π/3) f (x) − f (a)) 1
lim = lim = f 0 (a) = cos(π/3) = .
x→π/3 x − π/3 x→a x−a 2




III (1) Find f 0 (x) and f 00 (x) if f (x) = sin(x3 − 2).
Answer:

2
f 0 (x) = cos(x3 − 2) 3x2 , f 00 (x) = − sin(x3 − 2) 3x2 + cos(x3 − 2) 6x.






(2) Find g 0 (3) if h0 (9) = 17 and g(x) = h(x2 ).
Answer:
g 0 (x) = h0 (x2 )2x, g 0 (3) = h0 (9)6 = 102.




IV Find the constant c which makes g continuous on (−∞, ∞),
(
x2 if x < 4,
g(x) =
cx + 20 if x ≥ 4.
Answer: By the limit laws, g(x) is continuous at any x 6= 4.

lim g(x) = lim x2 = 16, lim g(x) = lim cx + 20 = 4c + 20.
x→4− x→4− x→4+ x→4+


The function g(x) is continuous when these are equal, i.e. when c = −1.

Answer:



V Consider the curve
y 2 + xy − x2 = 11.
(1) Find the equation of the tangent line to the curve at the point
P (2, 3).
Answer: Differentiate:
dy dy
2y +y+x − 2x = 0.
dx dx
Evaluate at the point P (2, 3):
dy dy
6 +3+2 − 4 = 0.
dx (x,y)=(2,3) dx (x,y)=(2,3)

Solve:
dy 1
= .
dx (x,y)=(2,3) 8

,This is the slope of the tangent line. The point P (2, 3) lies on the tangent line so the
equation of the tangent line is
(x − 2)
(y − 3) = .
8

d2 y
(2) Find at the point P (2, 3).
dx2
Answer: Differentiate again:
   
d2 y d2 y
2
dy dy dy
2 +y + + +x 2 − 2 = 0.
dx dx2 dx dx dx

Evaluate at (x, y) = (2, 3):
" # " #
d2 y d2 y
 2
1 1 1
2 +3 + + +2 − 2 = 0.
8 dx2 (x,y)=(2,3)
8 8 dx2 (x,y)=(2,3)


Solve:
1 2

1 1
d2 y −2 8
− 8
− 8
+2
= .
dx2 (x,y)=(2,3)
6+2

Answer:



VI Consider the function y = f (x) whose graph is shown below.
Match the expression in the left column with the correct correspond-
ing value in the right column.


f 0 (0) −6

f 0 (0.9) 0

f 0 (1) 3

f 0 (1.732) 0.6

, 2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2



Answer: For positive x, the slope of the tangent line is decreasing as x increases, so
f 0 (0) > f 0 (0.9) > f 0 (1) > f 0 (1.732). The only possibility is f 0 (0) = 3, f 0 (0.9) = 0.6,
f 0 (1) = 0, f 0 (1.732) = −6.

Answer:



VII Prove the product rule (f g)0 = f 0 g + f g 0 using the definition
of the derivative, high school algebra, and the appropriate limit laws.
Justify each step.
Answer: Let w(x) = f (x)g(x) and assume that f and g are differentiable.

w(x + h) − w(x)
w0 (x) = lim (1)
h→0 h
f (x + h)g(x + h) − f (x)g(x)
= lim (2)
h→0 h
f (x + h) − f (x) g(x + h) − g(x)
= lim g(x + h) + f (x) (3)
h→0 h h
f (x + h) − f (x) g(x + h) − g(x)
= lim lim g(x + h) + f (x) lim (4)
h→0 h h→0 h→0 h
= f 0 (x)g(x) + f (x)g 0 (x). (5)

Reasons:
(1) Definition of w0 (x).
(2) Definition of w(x).
(3) High school algebra.