Chapter 1 Worked Solutions
(c) We are given that f 0 (α) = 0. Use the proof from
(b) to establish that there is still a stationary point
at x = α.
1. Further Functions
1
f (x)
f 0 (x)
y0 = −
2
(f (x))
y=
Exercise 1A
Reciprocal and square root graphs
f 00 (x) (f (x)) − 2f (x) (f 0 (x))
(f (x))
4
Substitute x = α:
P3
(d)
2
2
y 00 = −
y 00 = −
y 2 = x4 − x2
1
4
=x 1− 2
x
r
1
∴ y = ±x2 1 − 2
x
r
2
4
(f (α))
00
f (α)
since f (α) 6= 0
=−
2,
(f (α))
Proving that it is still a turning point:
1
→ 1, so y ≈ x2 for
x2
sufficiently large values of x.
As x gets large,
f 00 (α) (f (α))
1−
P5
We are given that there is a turning point at x = α.
This means that there is a sign change of f 00 (x)
near x = α. This sign change transfers also to y 00
since the denominator is always positive, so there is
1
.
also a turning point at x = α for y =
f (x)
Proving that it has opposite concavity
0
(a) We are given that f (α) = 0 and f (α) 6= 0.
y=
p
f (x)
f 0 (x)
y0 = p
2 f (x)
When x = α, the denominator is zero and the
numerator is non-zero. Hence, there is a vertical
tangent at x = α.
It is important to note that the numerator is
0
non-zero because an indeterminate form of does
0
not guarantee a vertical asymptote. For example,
y = x2 has a square root graph of y = |x|, and this
does not have a vertical tangent. This is because if
√
2x
y = x2 , then y 0 = √ and substituting x = 0
2 x2
0
yields .
0
(b) We are given that f 0 (α) = 0.
1
f (x)
f 0 (x)
y0 = −
2
(f (x))
y=
When x = α, y 0 = 0 since f 0 (α) = 0 and f (α) 6= 0.
Hence, there is still a stationary point at x = α.
Recall that
y 00 = −
f 00 (α)
(f (α))
2
Since the denominator is always positive, the sign
of y 00 is always the negative of the sign of f 00 (x).
1
Hence, any turning point of y =
will have
f (x)
opposite concavity to y = f (x).
1
2
Chapter 1 Worked Solutions
Exercise 1B
Exercise 1C
Further reflections
Adding graphs
Q10
(a) Let g(x) = f (|x|).
P2
(a)
g(−x) = f (| − x|)
= f (|x|)
= g(x)
Hence, g(x) is even.
(b) Let g(x) = |f (x)|.
g(−x) = |f (−x)|
= | − f (x)|,
since f (x) is odd
= |f (x)|
= g(x)
Hence, g(x) is even.
P3
(a)
(b)
x−1
(x + 1) − 2
=
x+1
x+1
x+1
2
=
−
,
x+1 x+1
2
=1−
x+1
since
a+b
a b
= +
c
c
c
x+1
(x − 1) + 2
=
x−1
x+1
2
x−1
+
,
=
x−1 x+1
2
=1+
x+1
since
a+b
a b
= +
c
c
c
x
1
x
−1
+
=
+
x−1 1−x
x−1 x−1
x
1
=
−
x−1 x−1
x−1
=
x−1
=1
Chapter 1 Worked Solutions
Exercise 1D
Exercise 1E
Multiplying graphs
Inequalities
Q4
(b) Multplying both sides by (x + 1)2 :
2
≤4
x+1
2(x + 1) ≤ 4(x + 1)2
4(x + 1)2 − 2(x + 1) ≥ 0
(c)
4(x + 1)2 − 2(x + 1) ≥ 0
(x + 1) [4(x + 1) − 2] ≥ 0
(x + 1) [4x + 4 − 2] ≥ 0
(x + 1) [4x + 2] ≥ 0
2(x + 1) [2x + 1] ≥ 0
(x + 1)(2x + 1) ≥ 0,
since 2 > 0
3
4
Chapter 1 Worked Solutions
Exercise 1F
If f (x) is a self-inverse function, then applying f to f (x)
should return x.
a−x
1 + bx
a−x
a − 1+bx
=
a−x
1 + b 1+bx
Inverse functions
f (f (x)) = f
Q8
1 + bx
×
1 + bx
a(1 + bx) − (a − x)
(1 + bx) + b(a − x)
a + abx − a + x
=
1 + bx + ab − bx
abx + x
=
1 + ab
x(ab + 1)
=
1 + ab
=x
=
(c) This is because the graph of y = f −1 (x) is the
graph of y = f (x) reflected across y = x. So when a
point on y = x gets reflected across y = x, it of
course maps to itself. It is much like how when you
touch a mirror, your hand is ‘touching’ your
reflection in the mirror, but the point of contact is
the mirror itself.
P2
Q14
(a) f
(c) Swap x and y, then arrange:
x = y 2 − 2y
1
1
1
= + 1
a
a
a
1
= +a
a
= f (a), if a 6= 0
y 2 − 2y − x = 0
(d) Method #1: Quadratic formula
p
4 − 4(−x)
√ 2
2 ± 4 + 4x
=
2
√
2±2 1+x
=
√2
=1± 1+x
y=
2±
Method #2: Completing the square
y 2 − 2y = x
(b) If f (x) is not one-to-one, then the inverse is not a
function. For example, y = x2 is not one-to-one,
and the inverse is y 2 = x, which is not a function.
1
From the above result, we know that f (2) = f
2
and so on. This means that a single y-coordinate
can have more than one x-coordinate mapping to
it, which implies that f (x) is not one-to-one.
Hence, the inverse is not a function.
P3
1
1 + (−x)2
1
=
1 + x2
= f (x)
(a) f (−x) =
y 2 − 2y + 1 = 1 + x
(y − 1)2 = 1 + x
√
y−1=± 1+x
√
y =1± 1+x
P1
(b) By a similar reasoning to Problem 2, f (x) being
even means that a single y-coordinate may have
more than one x-coordinate mapping to it. In
particular, the x-coordinates mapping to it will be
negatives of each other, by symmetry of f (x).
Hence, f (x) is not one-to-one, and hence the
inverse is not a function.
Chapter Worked Solutions
P4
(b) First, swap x and y and re-arrange.
1
y
xy = y 2 − 1
x=y−
2
y − xy − 1 = 0
Exercise 1G
Parametric forms
P2
(a)
y
=
x
3t2
1+t3
3t
1+t3
2
1 + t3
3t
×
3
1+t
3t
=t
=
Use the quadratic formula:
p
x ± x2 − 4(−1)
y=
√ 2
x ± x2 + 4
=
2
Recall that the domain of f (x) is x > 0. Hence, the
range of f −1 (x) is y > 0. Hence, we must choose
the case with the plus sign to obtain this range.
√
x + x2 + 4
−1
∴ f (x) =
2
f −1 x2 + 4x + 5
p
= −2 + x2 + 4x + 5 − 1
p
= −2 + x2 + 4x + 4
p
= −2 + (x + 2)2
since x ≥ −2
=x
f −2 +
3t
= 0, t = 0 and hence y = 0.
1 + t3
Hence, the curve passes through the origin at
x = 0.
When x =
R19
a
We are given that x = ap, y = .
p
xy = ap ×
Proving that f f −1 (x) :
√
×x3
Now, we treat the x = 0 case separately, since it
was excluded from the above steps.
P5
(b) Proving that f −1 (f (x)):
= −2 + (x + 2),
3t
y
.
into x =
x
1 + t3
y
3
x
x=
y 3
1+
x
3x2 y
= 3
x + y3
3xy
1= 3
, for x 6= 0
x + y3
∴ x3 + y 3 = 3xy
(b) Substitute t =
a
= a2
p
Hence, it is an appropriate parametrisation of xy = a2 .
x−1
√
√
2
= −2 + x − 1 + 4 −2 + x − 1 + 5
√
√
= 4 − 4 x − 1 + (x − 1) − 8 + 4 x − 1 + 5
=x
It is important to note√for the proof of f −1 (f (x)),
we used the fact that x2 = x for x ≥ 0.
p
(x + 2)2 = x + 2, for x + 2 ≥ 0
= x + 2,
for x ≥ −2
5