A mistake in finding minimum value in quadratic function
Yue Kwok Choy
Question
1 − 4x − 4x 2 .
(a)
Find the maximum value of
(b)
Hence deduce the minimum value of
"Solution"
(a)
2−
3
.
4x + 4x − 1
2
(taken from New Way Mathematics Book 1 P.70, Practice 19 Teacher Edition)
2
2
2
2
 2

1
1
1
1
1
 2
1 1

1 − 4 x − 4x = −4 x + x −  = −4 x + x +   −   −  = −4  x +  −  = −4 x +  + 2
4
4 
2
2 
2

2 2



2
2
1

Since − 4 x +  ≤ 0 for all real values of
2

(b)
2−
3
3
= 2+
4x + 4x − 1
1 − 4x − 4 x 2
2
x, 1 − 4 x − 4 x 2 has the max. value of 2, when x = −
has the minimum value when the denominator of
1
.
2
3
has a
1 − 4 x − 4x 2
maximum value 2.
∴
The minimum value =
2+
3 7
=
2 2
Analysis
f(x) = 1 − 4 x − 4 x 2 of has a max. of 2 does not imply g(x) =
α=
Let the roots of f(x) = 0 be
−1− 2
,
2
β=
1
1
has a min. value of
.
2
1 − 4x − 4x
2
−1+ 2
.
2
y
It can be seen that from the diagram on the right,
(1)
f(x) ≥ 0
if
α≤x≤β.
(2)
f(x) < 0
if
x < α or x > β .
Since
g(x) =
2
1
,
f (x )
1
(1)
g(x) > 0
if
α<x<β.
(2)
g(x) < 0
if
x < α or x > β .
(3)
g(x) is undefined if x = α or x = β .
g(x) has only a local min. of
-2
O
-1
1
x
-1
1
1
when x = −
.
2
2
y
5
g(x) can be any negative value if x < α or x > β .
4
The graph of h(x) = 2 −
3
is shown on
4x + 4x − 1
3
2
2
the right diagram.
(1)
h(x) has a local min. of
(2)
h(x) < 2
7
when
2
if x < α or x > β .
x= −
1
.
2
1
-2
-1
O
1
2
x
1