Question 1:
(1). Compute
limx→0 (3x2 − 5)(x − 3)
lim (3x2 − 5)(x − 3) = (3(0)2 − 5)(0 − 3) = 15.
x→0
(2). Compute
x2 + 7x + 10
x+2
x2 + 7x + 10
(x + 2)(x + 5)
lim
= lim
x→−2
x→−2
x+2
x+2
= lim (x + 5)
limx→−2
x→−2
= (−2) + 5 = 3.
(3). Compute
sin 2x
5x
sin 2x 2x
sin 2x
= lim
·
lim
x→0 2x
x→0 5x
5x
sin 2x
2x
= lim
· lim
x→0 2x
x→0 5x
2
2
=1· = .
5
5
limx→0
(4). Compute
limx→∞
x2
(x − 5)(3 − x)
x2
x2
= lim
= −1.
x→∞ (x − 5)(3 − x)
x→∞ −x2 + 8x − 15
lim
1
2
Question 2:
The following function is not defined at a certain point. How should it be defined
in order to make it continuous at that point?
(a)
f (x) =
sin x
3x
(b)
f (x) =
x4 + 2x2 − 3
x+1
(a) The function f (x) is not defined at x = 0. However
sin x x
sin x
= lim
·
lim
x→0 x
x→0 3x
3x
x
sin x
· lim
= lim
x→0 3x
x→0 x
1
1
=1· = .
3
3
1
Therefore we should define f (0) = 3 to make f (x) being continuous at x = 0.
(b) The function f (x) is not defined at x = −1. However
x4 + 2x2 − 3
(x2 + 3)(x − 1)(x + 1)
= lim
x→−1
x→−1
x+1
x+1
2
= lim (x + 3)(x − 1)
lim
x→−1
= ((−1)2 + 3)(−1 − 1) = −8.
Therefore we should define f (−1) = −8 to make f (x) being continuous at x = −1.
3
Question 3:
Find the derivative of the following functions:
(a).
2x2 − 1
f (x) =
x+7
(2x2 − 1)0 (x + 7) − (2x2 − 1)(x + 7)0
f (x) =
(x + 7)2
(4x)(x + 7) − (2x2 − 1)(1)
=
(x + 7)2
2x2 + 28x + 1
=
.
(x + 7)2
0
(b).
f (x) =
sin x
tan x
0 0
sin x
sin x
=
f (x) =
tan x
sin x/ cos x
0
= (cos x) = − sin x.
0
(c)
f (x) = (x2 + 1)(1 − 3x − 4x3 )
f (x) = (x2 + 1)0 (1 − 3x − 4x3 ) + (x2 + 1)(1 − 3x − 4x3 )0
= (2x)(1 − 3x − 4x3 ) + (x2 + 1)(−3 − 12x2 )
= −20x4 − 21x2 + 2x − 3.
4
Question 4:
Find the equation of the tangent line to the function y = (x2 + 1)−1 at the point
(1, 21 ).
The slope of the tangent line at (1, 21 ) equals the derivative of y with respect to
x at x = 1. Since the derivative is
dy
(1)0 (x2 + 1) − (1)(x2 + 1)0
−2x
y 0 (x) =
=
= 2
,
2
2
dx
(x + 1)
(x + 1)2
we find the slope to be y 0 (1) = − 21 . Therefore the equation for the tangent line is
1
1
y − = − (x − 1).
2
2
1
(Or y = − 2 x + 1.)
5
Question 5:
Aircraft A is moving north with constant speed 10/second. After 4 seconds,
Aircraft B is sent to catch Aircraft A. The distance function of Aircraft B is 3t2
since then. Will Aircraft B catch Aircraft A within 6 seconds? Why?
(Hint: let f (t) be the distance function of A after B is sent and let g(t) be the
distance function of B. B catches A when g(t) = f (t).)
Let f (t) be the distance function of A after B is sent. Then f (t) = 10t+4(10) =
10t + 40.
Let g(t) be the distance function of B. Then g(t) = 3t2 .
B catches A when they travels the same amount of distance. Mathematically,
this means g(t) − f (t) = 0. Now let
h(t) = g(t) − f (t) = 3t2 − (10t + 40) = 3t2 − 10t − 40.
We want to check whether there is a root of h(t) for t between 0 and 6. Since
h(t) is continuous, we can try to use the Intermediate Value Theorem. We have
h(0) = −40 and h(6) = 8. Thus the Intermediate Value Theorem tells us that
there must be a root in the interval [0, 6]. This implies that B catches A within 6
seconds.