Section Check In – 1.02 Algebra and Functions
Questions
1.
Use the factor theorem to show that (2 x  1) is a factor of 2 x3  3x2  7 x  3 .
2.
Simplify
3.*
Solve the inequality 3x  4  2 x  1 , giving your answer using set notation.
4.*
Simplify
5.
(a) Express each of x2  6 x  27 and 1  6 x  x 2 in completed square form.
 12
12  6  3
1
 27 2 .
x3  x 2  9 x  9
.
( x 2  2 x  3)(3x 2  8 x  3)
(b) Sketch the curves y  x 2  6 x  27 and y  1  6 x  x 2 on the same axes, and show
that the shortest distance between the two curves is 8 .
6.*
The functions f and g are defined by
f ( x)  x 2  2 x with domain
,
g(x)  x with domain x  0 .
(a)
(b)
(c)
(d)
Show that the range of f is {f ( x) : f ( x)  1} .
Explain why f has no inverse.
Explain why the function gf does not exist.
Solve the equation fg( x)  15 .
7.
Show that the straight line y  x  k meets the curve x 2  3xy  y 2  5 in two distinct points
for all values of the constant k .
8.*
(a) Express
4 x
4 x
in partial fractions and hence expand
in ascending
2
1  x  2x
1  x  2x2
powers of x up to and including the term in x 2 .
(b) Substitute a suitably small value of x in both the original rational expression and in your
expansion. Comment on the two results obtained.
9.
An object is projected vertically upwards from ground level and its height, H metres, after t
seconds is given by the formula
H  60t  5t 2 .
Show that the object is above a height of 90 metres for approximately 8.5 seconds.
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10.* The design of the slide in a playground is based on two curves
y  f ( x) for 0  x  6 and y  g( x) for 6  x  20 .
Units are metres and ground level is represented by y  0 .
The curve y  f ( x) is the result of transforming the curve y 
 4 
1
by the translation  
2
x
0
followed by a stretch in the y - direction with scale factor 160 .
It is given that g( x ) 
k
 c for constants k and c . The end of the slide (where x  20 )
x
must be at a height of 0.2 metres above the ground.
Find formulae for f ( x) and g( x) .
Extension
A self-inverse function is one for which f ( x)  f 1 ( x) . By applying f to both sides, we see that this
is equivalent to f 2 ( x)  x .
(a)
Verify that the function defined by f ( x)  4  x for all real values of x is self-inverse.
(b)
Find conditions on the constants a and b so that the function defined by f ( x)  ax  b is
self-inverse.
(c)
Similarly investigate f ( x) 
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c
px  q
and f ( x) 
.
dx  e
rx  s
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Worked solutions
1.
Let f ( x)  2 x3  3 x 2  7 x  3
f ( 12 )  2  18  3  14  7  12  3  14  43  3 12  3  0
Since f ( 12 )  0 , by factor theorem x 
1
1
2
or 2x  1 is a factor
1
12  6  3 2  27 2  4  3 
2.
Simplifying,
3.
Squaring both sides, (3 x  4) 2  (2 x  1) 2
6
3
 93  2 3 
6 3
3 3  3
3
Expanding and simplifying, 5x2  28x  15  0
Factorising, (5 x  3)( x  5)  0 and so x  53 and x  5 are critical values
Checking values x  0 and x  6 , for example, satisfies the inequality and so the solution
consists of values to the left of 53 and to the right of 5
Solution is {x : x  53}  {x : x  5}
Alternative solution:
Sketching y  3x  4 and y  2 x  1
y
4
1
x
1
−
2
4
3
Points of intersection given by 2 x  1  (3x  4)
and by 2x  1  3x  4 , i.e. when x  53 and x  5
From diagram, y  3x  4 is ‘above’ y  2 x  1 for the values {x : x  53}  {x : x  5}
4.
By inspection or use of factor theorem, ( x  1) is factor of the numerator and
x3  x 2  9 x  9  ( x  1)( x 2  9)
Expression is
( x  1)( x 2  9)
( x 2  2 x  3)(3x 2  8 x  3)
Factorising the quadratic expressions,
( x  1)( x  3)( x  3)
( x  3)( x  1)(3x  1)( x  3)
Cancelling factors common to numerator and denominator, expression is
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1
3x  1
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5.
(a) x 2  6 x  27  ( x  3) 2  9  27  ( x  3) 2  18
1  6 x  x 2  1  ( x  3) 2  9  10  ( x  3) 2
(b) y  ( x  3) 2  18 has minimum at (3, 18)
y  10  ( x  3) 2 has maximum at (3, 10)
y
27
1
x
Shortest distance between curves is the distance between their stationary points.
Shortest distance is 18  10  8
6.
(a) Completing the square, f ( x)  x 2  2 x  ( x  1) 2  1
For all values of x , ( x  1) 2  0 and therefore f ( x)  1 for all x 
(b) To have an inverse, f must be one-one so that f 1 ( x) always has a unique value; but,
for example, f ( 3) and f (1) are both equal to 3 and so f is not one-one and therefore
has no inverse.
(c) For the composed function gf to exist, the range of f must be part of, or the same as,
the domain of g so that gf ( x) can be found for all values of x in the domain of f ; but
the range of f includes some negative values which are not included in the domain of
g ; for example, gf ( 12 )  g( 34 ) and this cannot be found
(d) The equation fg( x)  15 leads to f ( x )  15 and therefore ( x ) 2  2 x  15 or
x  2 x  15  0
Factorising (or using a substitution u 
Solving
But
7.
x  5 or
x ), ( x  5)( x  3)  0
x 3
x  3 , giving x  9 as the only solution
x cannot be negative and so
Substituting y  x  k in the equation of the curve, x 2  3 x( x  k )  ( x  k ) 2  5
Expanding, x2  3x2  3kx  x2  2kx  k 2  5  0
Simplifying, 5x2  5kx  k 2  5  0
Discriminant, b 2  4ac  (5k ) 2  4  5  (k 2  5)  25k 2  20k 2  100  5k 2  100
For all values of k , k 2  0 and therefore 5k 2  100  0
Since the discriminant is positive, the quadratic equation has two distinct roots and so the
line meets the curve in two distinct points.
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8.
4 x
4 x
A
B



2
(1  2 x)(1  x) 1  2 x 1  x
1  x  2x
Multiplying by (1  2 x)(1  x) gives 4  x  A(1  x)  B(1  2 x)
Substituting x  1 gives 3  A  0  B  3 and therefore B  1
Substituting x   12 gives 4 12  A 1 12  B  0 and therefore A  3
(a) Let
Expression is 3(1  2 x) 1  (1  x) 1
Using binomial expansions, expression is 3(1  2 x  (2 x) 2  ...)  (1  x  x 2  ...)
Simplifying, expansion is 4  5x  13x2
(b) Expansion is valid only if  12  x  12 so value chosen must be between these values
Choosing x  0.01 to substitute: using calculator, value of
4 x
is 3.951277 481
1  x  2x2
and value of 4  5x  13x2 is 3.9513
The values agree correct to 5 significant figures; agreement would be closer if more
terms had been used in the expansion or if a smaller value of x had been chosen (or
both); choosing a particular value in this way does not prove that the expansion is
correct but no obvious error has been revealed
9.
Times when object is at a height of 90 metres found by substituting H  90
Equation is 90  60t  5t 2 which simplifies to t 2  12t  18  0
Using quadratic formula, t  1.757... or t  10.242...
Object at height of 90 metres after 1.757... seconds (going up) and after 10.242... seconds
(going down)
Object above this height for 10.242...  1.757...  8.485... seconds, i.e. for 8.49 seconds (to 3
significant figures)
10.
 4 
1
1
 applied to the curve y  2 gives the curve y 
x
( x  4) 2
0
1
160
The stretch applied to y 
gives the curve y 
2
( x  4)
( x  4) 2
The two sections of the slide must meet when x  6 so that g(6)  f (6) , giving the
k
160
equation  c 
6
100
At the end of the slide when x  20 , the height of the slide is 0.2 m , i.e. g(20)  0.2 giving
k
 c  0.2
a second equation
20
Solving the simultaneous equations for k and c gives k  12 and c  0.4
12
160
 0.4
Hence f ( x) 
and g( x) 
2
x
( x  4)
A translation of 
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Extension
(a)
f ( x)  4  x so that f 2 ( x)  ff ( x)  f (4  x)  4  (4  x)  4  4  x  x
(b)
Since f 2 ( x)  x , this function f is self-inverse
Now consider f ( x)  ax  b
If f 2 ( x)  x , then f ( ax  b)  x and so a (ax  b)  b  x
Rearranging gives (a 2  1) x  (a  1)b  0
For this to be true for all values of x , a2  1  0 and (a  1)b  0
Therefore either a  1 and b  0 or a  1 and b can take any value
The first option is just the trivial case f ( x)  x
The second option is f ( x)   x  b , i.e. the function f defined by f ( x)  b  x is a
self-inverse function for all values of the constant b
(c)
For the function f ( x) 
c
, applying f 2 ( x)  x and simplifying leads to
dx  e
dex2  e2 x  ce  0
This means de  0 and e2  0 and ce  0 giving only e  0
c
Hence the function f defined by f ( x ) 
is a self-inverse function for all values of the
dx
non-zero constants c and d
A similar approach in the final case shows the function f defined by f ( x) 
inverse function provided that p  s  0 and r  0
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px  q
is a selfrx  s
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