FUNCTIONS
Functions worksheet 1
1.
2.
If y = x3 – 3x and z = 2y – 1 :
(a)
Find y when x = 2
(b)
Find z when y = 5
(c)
Find z when x = 3
If f(x) = x2 – 1 ; g(x) = x2 + 1 ; and h(x) = 1/(x + 1) :
(a)
Find values for: f(5) ; g(11); h(4.2)
(b)
Find expressions (reasonably simplified) for
(c)
(d)
3.
f(x – 1) ;
f(2x + 3) ;
f(4x)
g(x – 1) ;
g(2x + 3) ;
g(4x)
h(x – 1) ;
h(2x + 3) ;
h(4x)
Find expressions (reasonably simplified) for
f(f(x)) ;
g(f(x)) ;
h(f(x))
f(g(x)) ;
g(g(x)) ;
h(g(x))
f(h(x)) ;
g(h(x)) ;
h(h(x))
For any of these functions, note where there are restrictions on the
domain.
If r(x) = 1/x , s(x) = x2, find expressions and domains for
r (r (x))
s (r (x))
r (s (x))
s (s (x))
r (√x)
s (√x)
√ r(x)
√ s(x)
4.
Identify the natural domain and the range for each of the functions
(a)
(d)

f (x)  3x x – 2
m(x) 
x2
x 1
(b) g(x) 
3x
x2
(c)
h(x) 
3x
x 2
(e) p(x) 
x2
x 1
(f)
q(x) 
x 2
x 1


r(x)  x 1  x  3
(h)
s(x)  x 1  x  3
(i)
t(x)  (x 1)(x
 3)
(j)
u(x) 
 x 1  3  x

(k)
v(x)  (x 1)(3  x)

(l)

(g)

w(x) 
Functions – Conic Sections
5.
6.
Sketch a graph of each of the following:

(a)
x2 + y2 = 16
(b)
x2 + 4y2 = 16
(c)
x2 – 4y2 = 16
(d)
4y2 – x2 = 16
(e)
x2/9 + y2/4 = 1
(f)
x2/16 – y2/9 = 1
(g)
y2/8 – x2/4 = 1
(h)
x2/8 – y2/9 = 0
Sketch a graph of each of the following:
(a)
(x – 3)2 + (y + 2)2 = 16
(b)
(x – 1)2 + 4(y + 1)2 = 16
(c)
4(x – 2)2 – (y + 3)2 = 4
(d)
(y – 2)2 – 3x2 = 9
(e)
x2 + y2 – 2x – 6y = –1
(f)
3x2 – y2 + 6x – 4y = 10
(g)
4x2 + y2 – 16x + 2y + 1 = 0
(h)
–3x2 + y2 + 6x – 4y = 8
1
x 1  3  x
Inverse Functions
7.
Each of the following functions is one-to-one. In each case find a formula for
the inverse function as a function of x. You should check also [for (a) to (f)] that the
graphs of each function and its inverse are reflections of each other in the line y = x,
(b)
y = 2x3 – 5
1
(x ≠ –1)
x 1
(d)
y
x4
(x ≠ 3)
x3
2x 1
(x ≠ –2)
x2
(f)
y
x4
(x ≠ 1/3)
3x – 1
y
ax  b
(x ≠ –d/c)
cx  d
(a)
y = 2x – 5
(c)
y
(e)
y

(g)

y = mx + c
(h)


 a restricted domain is specified, so that the
8.
For each of the following functions
function is one-to-one on that domain. Determine the corresponding range; then find
a formula for the inverse function as a function of x, noting both the domain and
range of the inverse function.
(a)
(b)
(c)
(d)
(e)
(f)

y = x2 – 3 ; x ≥ 0
y = x2 + 4x – 5; x ≥ –2
1
y 2
; x≥0
x 1
y  x 2 1 ; x ≥ 0
y  4  2x 2 ; 0 ≤ x ≤ √2
y  8  2x  x 2 1 ≤ x ≤ 4


Functions
Worksheet 1 Answers
1.
(a)
2
(b)
9
2.
(a)
24; 122; 0.19231
(b)
[f]
(c)
35
(x – 1)2 – 1 = x2 – 2x
(2x + 3)2 – 1 = 4x2 + 12x + 8
(4x)2 – 1 = 16x2 – 1
[g]
(x – 1)2 + 1 = x2 – 2x + 2
(2x + 3)2 + 1 = 4x2 + 12x + 10
(4x)2 + 1 = 16x2 + 1
[h]
(c)
1/x ; 1/(2x + 4) ; 1/(4x + 1)
f(f(x)) = (x2 – 1)2 – 1 = x4 – 2x2
g(f(x)) = (x2 – 1)2 + 1 = x4 – 2x2 + 2
h(f(x)) = 1/[(x2 – 1) + 1] = 1/x2
f(g(x)) = (x2 + 1)2 – 1 = x4 + 2x2
g(g(x)) = (x2 + 1)2 + 1 = x4 + 2x2 + 2
h(g(x)) = 1/[(x2 + 1) + 1] = 1/(x2 + 2)
 1 2
x 2  2x
f (h(x))  
 1  
x  1
(x  1) 2

 1 2
x 2  2x  2
g(h(x))  

1


x  1
(x  1) 2
h(h(x)) 

(d)

1
1
x 1
1

x 1
[N.B. last entry in (d)]
x2
h(x) : x ≠ –1
In (b): all are defined for all x except:
h(x – 1) = 1/x : x ≠ 0 [so x – 1 ≠ –1]
h(2x + 3) = 1/(2x + 4) : x ≠ –2 [2x + 3 ≠ –1]
h(4x) = 1/(4x + 1) : x ≠ –1/4 [4x ≠ –1]
In (c): h(f(x)): x ≠ 0 [as f(0) = –1]
h(g(x) is defined for all x, as g(x) ≠ –1.
f(h(x)) and g(h(x)) : x ≠ –1
h(h(x)) : x ≠ –1 AND x ≠ –2 (as h(–2) = –1)
3.
r(r(x)) = x (but x ≠ 0)
s(r(x)) = 1/x2 (x ≠ 0)
r(s(x)) = 1/x2 (x ≠ 0)
s(s(x)) = x4 (all x)
r(√x) = 1/√x (x > 0)
s(√x) = x (x ≥ 0)
√ r(x) = 1/√x (x > 0)
√ s(x) = | x | (all x)
4.
(a) domain x ≥ 2; range y ≥ 0
(b) domain x ≥ –2; range all y
(c) domain x ≥ 2; range y ≥ 6√2 [minimum when x = 4]
(d) domain x ≠ 1; range y ≠ 1
(e) domain x ≤ –2 or x ≥ 1; range 0 ≤ y < 1 or y > 1
(f) domain x < 1 or x ≥ 2; range 0 ≤ y < 1 or y > 1
(g) domain x ≥ 3; range y ≥ √2
(h) domain x ≥ 3; range y ≥ 0
(i) domain: x ≤ 1 or x ≥ 3; range y ≥ 0
(j) domain 1 ≤ x ≤ 3; range √2 ≤ y ≤ 2
(k) domain 1 ≤ x ≤ 3; range 0 ≤ y ≤ 1
(l) domain 1 ≤ x < 2 or 2 < x ≤ 3; range y ≤ –1/√2 or y ≥ 1/√2
5.
(a)
(b)
(c)
(d)
(e)
(f)
6.
(g)
(h)
(a)
(b)
(d)
(e)
(c)
(x – 1)2 + (y – 3)2 = 9
(f)
3(x + 1)2 – (y + 2)2 = 9
(g)
4(x – 2)2 + (y + 1)2 = 16
[Upper and lower sections should join]
7.
8.
(h)
–3(x – 1)2 + (y – 2)2 = 9
(a)
y = (x + 5)/2
(c)
y = 1/x – 1 (x ≠ 0)
(e)
y = – (2x + 1)/(x – 2) (x ≠ 2)
(f)
y = (x + 4)/(3x – 1) (x ≠ 1/3) [this function is self-inverse]
(g)
y = (x – c)/m (h)
(a)
y = x2 – 3 ; x ≥ 0 : range y ≥ –3
(b)
y = 3√((x + 5)/2)
(d)
y = (3x + 4)/(x – 1) (x ≠ 1)
y = – (dx – b)/(cx – a) (x ≠ a/c)
inverse y = √(x + 3) ; x ≥ –3 ; y ≥ 0
(b)
(c)
y = x2 + 4x – 5; x ≥ –2
inverse y = –2 + √(x + 9) ; x ≥ –9 ; y ≥ –2
y
1
; x ≥ 0 : range 0 < y ≤ 1
x 1
2
inverse y = √(1/x – 1) ; 0 < x ≤ 1 ; y ≥ 0

(d)
y  x 2 1 ; x ≥ 0 : y ≥ 1
inverse y = √(x2 – 1) ; x ≥ 1 ; y ≥ 0

(e)
y  4  2x 2 ; 0 ≤ x ≤ √2 : 0 ≤ y ≤ 4
inverse y = √[(4 – x2)/2] ; 0 ≤ x ≤ 4 ; 0 ≤ y ≤ √2

(f)
y  8  2x  x 2 1 ≤ x ≤ 4 : 0 ≤ y ≤ 3
inverse y = 1 + √(9 – x2) ; 0 ≤ x ≤ 3 ; 1 ≤ y ≤ 4
