MATH10330 2013
Test 1: Solutions
Attempt all questions for full marks. Write your answers in the answer
book provided.
1. (a) In a single diagram draw the graphs y = sin(x) and
y = sin(x − π/2), labelling both graphs and indicating clearly
the roots of both functions.
(b) In a single diagram draw the graphs y = cos(x) and y = 3 cos(x),
labelling both graphs and indicating clearly the roots of both
functions.
Solution:
(a) sin(x − π/2) is the dashed curve:
(b) 3 cos(x) is the dashed curve:
2. Evaluate the following limits (where they exist), showing your work
or explaining your reasoning:
x2 − 6x + 9
x→3
x2 − 9
x2 − 6x + 9
(b) lim
x→−3
x2 + 9
sin(3x)
(c) lim
x→0 sin(4x)
sin(3x)
(d) lim
x→0 cos(4x)
(a) lim
Solution:
(a) Since
x2 − 6x + 9
(x − 3)2
x−3
=
=
2
x −9
(x − 3)(x + 3)
x+3
we have
x2 − 6x + 9
x−3
0
= lim
= =0
2
x→3
x→3
x −9
x+3
6
lim
(b) On the other hand
x2 − 6x + 9
36
lim
=
=2
2
x→−3
x +9
18
(c) Recall that
sin(ax)
ax
= 1 = lim
x→0
x→0 sin(ax)
ax
lim
for any a.
Thus
sin(3x)
lim
= lim
x→0 sin(4x)
x→0
3 sin(3x) 4x
4 3x sin(4x)
=
3
sin(3x)
4x
3
3
lim
lim
= ·1·1= .
4 x→0 3x x→0 sin(4x)
4
4
(d) Since f (x) = sin(3x)/ cos(4x) is a continuous function and x = 0
is in its domain
lim f (x) = f (0) =
x→0
sin(0)
0
= = 0.
cos(0)
1
3. Consider the function

 1 + sin(x), x ≤ 0
2x ,
0<x<2
f (x) =

3x − 2,
x≥2
(a) Is this function continuous? Explain your answer.
(b) Sketch the graph of this function.
Solution:
(a) The function is continuous:
For x < 0 it coincides with the continuous function 1 + sin(x).
If 0 < x < 2 it coincides with the exponential (hence continuous) function 2x . If x > 2 it coincides with the linear (hence
continuous) function 3x − 2.
It remains to demonstrate continuity at x = 0 and at x = /2:
Now
lim f (x) = lim (1 + sin(x)) = 1 + 0 = 1 = f (0)
x→0−
x→0−
while
lim f (x) = lim 2x = 20 = 1 = f (0)
x→0+
x→0+
Thus limx→0 f (x) exists and equals f (0) as required.
Similarly
lim f (x) = lim 2x = 22 = 4 = f (2)
x→2−
x→2−
and
lim f (x) = lim (3x − 2) = 3 · 2 − 2 = 4 = f (2)
x→2+
x→2+
so that f is continuous at 2 also.
(b) Here is the graph of f :
4. Use the Intermediate Value Theorem to show that there is some real
number whose square is exactly 2 less than its cube.
Solution: We wish to show that there is a real number satisfying
x2 = x3 − 2; i.e. x3 − x2 − 2 = 0. Thus we wish to show that
f (x) = x3 − x2 − 2 has a root.
Now f is polynomial, hence continuous on R. Also f (1) = −2 < 0
while f (2) = 2 > 0. By the Intermediate Value Theorem there is
some x ∈ (1, 2) satisfying f (x) = 0, as required.
5. Differentiate the function f (x) = cos(x) from first principles, showing
all of your work. (You may use the limits limt→0 sin(t)/t = 1 and
limt→0 (cos(t) − 1)/t = 0, as well as the identity
cos(a + b) = cos(a) cos(b) − sin(a) sin(b).)
Solution:
cos(x + h) − cos(x)
d
(cos(x)) = lim
h→0
dx
h
cos(x) cos(h) − sin(x) sin(h) − cos(x)
= lim
h→0
h
cos(h) − 1
sin(h)
= lim cos(x)
− sin(x)
h→0
h
h
cos(h) − 1
sin(h)
= cos(x) lim
− sin(x) lim
h→0
h→0
h
h
= cos(x) · 0 − sin(x) · 1 = − sin(x).
6. Differentiate the following functions of t:
√
√
1
(a) t + t2 (b) √
(c)
t tan(t) Solution: (a)
4
t
d √
1
( t + t2 ) = √ + 2t.
dt
2 t
(b)
d
dt
(c)
1
√
4
t
=
d −1/4
1
(t
) = − t−5/4 .
dt
4
√
d√
tan(t)
t tan(t) = t sec2 (t) + √ .
dt
2 t