Tutorial 1: SOLUTIONS
Functions
1. We’re given the piecewise function
(√
x≥1
3
x < 1.
√
√
Then g(3) = 3 + 1 = 2, g(−1) = 3 and g(π) = π + 1. To obtain g(t2 − 1)
is a bit trickier since we need to take the cases when t2 − 1 ≥ 1 and t2 − 1 < 1
separately.
g(x) =
x+1
If t satisfies t2 − 1 < 1, then g(t2 − 1) = 3 by definition of the piecewise √
function.
2
This corresponds to t satisfying t − 2 < 0 which may be solved to get − 2 < t <
√
2.
√
On the other hand, if t satisfies t2 − 1 ≥ 1, then g(t2 − 1) = t2 − 1 + 1 = t by
2
definition of √
the piecewise
√ function. Again we can solve the inequality t − 1 ≥ 1
to get t ≤ − 2 ∪ t ≥ 2. We can represent our solution as a piecewise function
(
√
√
3
− 2<t< 2
2
g(t − 1) =
t
otherwise.
2. We wish to find the natural domain of the following functions:
√
(a) f (x) = x2 + x − 6. The square root is well defined so long as what’s inside
the square root is non-negative, so the domain is defined by all x satisfying
the inequality
=⇒
=⇒
x2 + x − 6 ≥ 0
(x − 2)(x + 3) ≥ 0
x ≤ −3
and
x ≥ 2.
Hence the domain is
D(f ) = (−∞, −3] ∪ [2, ∞).
1
. In this case, all x such that the denominator is zero are
1 − sin x
NOT in the domain of f . Solving
(b) f (x) =
sin x = 1
=⇒
x = π/2 ± 2nπ,
n ∈ N.
Formally then, we write the domain as
D(f ) = {x ∈ R : x 6= π/2 ± 2nπ, n ∈ N}
or simply all x except x = π/2 ± 2nπ where n ∈ N.
1
3
. Again, all x such that the denominator vanishes are NOT
2 − cos x
in the domain. However, in this case, the equation
(c) f (x) =
cos x = 2
has no solutions since cos x ranges between −1 and 1, i.e. there is no x such
that cos x = 2. Hence the domain is the entire real line:
D(f ) = (−∞, ∞).
3. We are required to express the following functions in piecewise form without using
absolute values:
(a) f (x) = |x| − 7x + 1. The presence of the |x| term implies we take the x < 0
and x ≥ 0 cases separately.
x<0
x≥0
=⇒
|x| = −x
(by definition),
=⇒ f (x) = −x − 7x + 1 = −8x + 1.
=⇒
|x| = x
(by definition),
=⇒ f (x) = x − 7x + 1 = −6x + 1.
Hence, written as a piecewise function, we obtain
(
−8x + 1
x<0
f (x) =
−6x + 1
x ≥ 0.
(b) f (x) = 4|x − 2| − |x + 2|. The presence of both a |x − 2| and a |x + 2| term
implies we take three separate cases: when x < −2, when −2 ≤ x < 2 and
when x ≥ 2.
x < −2
−2 ≤ x < 2
x≥2
=⇒
x − 2 < 0, x + 2 < 0
=⇒ |x − 2| = −(x − 2), |x + 2| = −(x + 2)
=⇒ f (x) = −4(x − 2) + (x + 2) = −3x + 10.
=⇒
x − 2 < 0, x + 2 ≥ 0
=⇒ |x − 2| = −(x − 2), |x + 2| = x + 2
=⇒ f (x) = −4(x − 2) − (x + 2) = −5x + 6.
=⇒
x − 2 ≥ 0, x + 2 > 0
=⇒ |x − 2| = x − 2, |x + 2| = x + 2
=⇒ f (x) = 4(x − 2) − (x + 2) = 3x − 10.
Hence written as a piecewise function, we get


x < −2
−3x + 10
f (x) = −5x + 6
−2 ≤ x < 2


3x − 10
x ≥ 2.
2
4. We are asked to compare the domains of the functions
f (x) =
(x + 2)(x2 − 1)
(x + 2)(x − 1)
and
g(x) = x + 1
and to plot them on separate graphs. The first point to note is that f (x) is clearly
not well-defined at x = −2 and x = 1 since the denominator is zero at these points.
However, when x 6= −2 and x 6= 1, we can cancel the (x + 2) and (x − 1) factors in
the definition of f (x) to get f (x) = x + 1, i.e., f (x) = g(x) for all x except x = −2
and x = 1 and f (x) is NOT defined at these points. Hence the domain of f (x) is
all x ∈ R except these two points while the domain of g(x) is the entire real line.
Their graphs look like
6
!2
5
5
4
4
3
3
2
2
1
1
!1
1
2
3
4
y = g(x)
6
y = f (x)
!2
5
!1
!1
1
2
3
4
5
!1
5. We are given the graph of a function y = f (x) which is shown in red in the figure
below.
2
1
−4 −3 −2 −1
1
2
3
−1
We are asked to plot the graphs of the following:
3
4
(a) y = f (x) − 1. This is obtained by moving the graph of f (x) down by one
unit. This corresponds to the blue graph in the figure above.
(b) y = f (x − 1). This is obtained by moving the graph of f (x) to the right by
one unit. This corresponds to the green graph in the figure above.
(c) y = 21 f (x). This is obtained by compressing the graph of f (x) by a factor of 2
in the y-direction. This corresponds to the purple graph in the figure above.
(d) y = f (− 21 x). The graph of y = f ( 21 x) would be obtained by stretching the
graph of f (x) by a factor of 2 in the x-direction and the graph of y = f (− 12 x)
is obtained by reflecting the graph of y = f ( 12 x) through the y-axis. This
corresponds to the yellow graph in the figure above.
√
x
6. We are given the functions f (x) = 1+x
x − 3, we wish to find the
2 and g(x) =
domain of f /g. For (f /g)(x) to be well defined, x must be in the domain of f and
the domain of g and g(x) cannot be zero. Formally we can write this as
D(f /g) = D(f ) ∩ D(g)/{x ∈ R : g(x) = 0},
but don’t worry if you find this notation hard to read, you need not use it if you
don’t like it. Anyway the domain of f is the entire real line since the denominator
cannot be zero while the domain of g will be all x values such that what’s inside
the square root is non-negative, i.e.
D(f ) = (−∞, ∞),
D(g) = [3, ∞)
and hence the intersection is
D(f ) ∩ D(g) = [3, ∞).
But we must exclude any values for which g(x) = 0 since that would give us a zero
in the denominator of (f /g)(x). Clearly g(x) = 0 whenever x = 3, excluding this
value means we change the square bracket to a round bracket, hence
D(f /g) = (3, ∞).
√
√
7. For f (x) = x − 3 and g(x) = x2 + 3, we must find f ◦ g and g ◦ f and the
domains of the compositions. First,
qp
p
(f ◦ g)(x) = f (g(x)) = f ( x2 + 3) =
x2 + 3 − 3.
By definition, the domain of f ◦ g is all x ∈ D(g) for which g(x) is in the domain of
f . The condition that x ∈ D(g) is automatically satisfied here since the domain of
g is the entire real line (since what’s inside the square root is always positive). We
4
also require that g(x) ∈ D(f ) where D(f ) = [3, ∞). This is equivalent to requiring
that g(x) ≥ 3, i.e.,
p
x2 + 3 ≥ 3
x2 + 3 ≥ 9
=⇒
x2 − 6 ≥ 0
√
√
x ≤ − 6, x ≥ 6.
=⇒
=⇒
Hence the domain is
√
√
D(f ◦ g) = (−∞, − 6] ∪ [ 6, ∞).
Second, the composition g ◦ f is
q√
√
√
√
(g ◦ f )(x) = g(f (x)) = g( x − 3) = ( x − 3)2 + 3 = x − 3 + 3 = x.
The domain of g ◦ f is all x ∈ D(f ) for which f (x) ∈ D(g). Since the domain of g
is the entire real line, the condition f (x) ∈ D(g) is automatically satisfied so long
as x ∈ D(f ). Hence the domain of g ◦ f is the same as the domain of f :
D(g ◦ f ) = [3, ∞).
8. We are asked to sketch the graphs of the following functions by making an appropriate transformation of a basic function:
√
√
(a) y = 2 + x − 3. This is obtained by shifting the graph of y = x two units
up and three units to the right, as seen in the figure below.
y =2+
4
√
x−3
3
2
1
y=
√
2
x
4
6
8
10
(b) y = 12 cos(3x). This is obtained by compressing the graph of y = cos x by a
factor of 2 vertically and a factor of 3 horizontally. The graph is given below:
5
!6
!4
1.0
y = cos(x)
0.5
y=
!2
2
4
1
2
cos(3x)
6
!0.5
!1.0
9. For the given functions, we wish to apply the horizontal line test to determine
whether or not the functions have an inverse over the domain of f . The solution
is sketched in the graphs below:
y = 2x − 1
4
y=
1.5
√
2.0
x
y = |x − 1|
1.5
2
1.0
1.0
!1
1
2
3
0.5
0.5
Inverse
!2
Inverse
0.5
2
1.0
1.5
4
y = x3
2.0
2.5
No Inverse
3.0
!1
1
1.0
y = x2 − x − 2
3
2
3
y = cos x
0.5
2
1
1
!6
!2
!1
1
!2
!1
1
2
!2
2
4
6
3
!0.5
!1
!2
!4
2
!1
Inverse
No Inverse
!2
!1.0
No Inverse
10. For the given functions, we wish to find f −1 (x) and state its domain:
(a) f (x) = (x + 2)4 ,
x ≥ 0. We set y = (x + 2)4 and solve for x in terms of y.
=⇒
=⇒
6
y = (x + 2)4
√
4
y =x+2
√
x= 4y−2
√
Interchanging x and y gives y = 4 x − 2 which defines the inverse function,
i.e.
√
f −1 (x) = 4 x − 2.
Now the domain of f −1 is equal to the range of f . As x varies between 0 and
∞, f (x) ranges from 16 to ∞. Hence
D(f −1 ) = R(f ) = [16, ∞).
√
(b) f (x) = − 2 − 3x. We follow the same method:
√
y = − 2 − 3x
y 2 = 2 − 3x
=⇒
3x = 2 − y 2
=⇒
=⇒
x=
2 − y2
.
3
Again, we interchange x and y which defines the inverse
f −1 (x) =
2 − x2
.
3
The domain of f in this case is (−∞, 2/3], as x varies over this domain f (x)
ranges over (−∞, 0]. Hence
D(f −1 ) = R(f ) = (−∞, 0].
11. We wish to sketch the parametric curve (x = 2t − 1, y = t + 1) and indicate its
orientation. We take a series of t values and compute the corresponding coordinates:
t = −1 : (x = −3, y = 0),
t = 2 : (x = 3, y = 3),
t = 0 : (x = −1, y = 1),
t = 3 : (x = 5, y = 4),
etc.
Plotting these, we get
5
t=3
4
t=2
3
2
t=1
t=0
1
t = −1
!4
!2
2
!1
7
4
6
t = 1 : (x = 1, y = 2)