CIE/GCE/AS/Math/P1/19/Nov/13/Q#2
Question
The function g is defined by 𝑔(𝑥 ) = 𝑥 2 − 6𝑥 + 7 for 𝑥 > 4. By first completing the
square, find an expression for 𝑔−1 (𝑥)_and state the domain of 𝑔−1 .
Solution
We are given that;
𝑔(𝑥 ) = 𝑥 2 − 6𝑥 + 7
We use method of “completing square” to obtain the desired form. We complete the
square for the terms which involve 𝑥.
(𝑥 2 − 6𝑥 ) + 7
We have the algebraic formula;
(𝑎 − 𝑏)2 = (𝑎)2 − 2(𝑎)(𝑏) + (𝑏)2
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2
For the given case we can compare the given terms with the formula as below;
𝑎2 = 𝑥 2
2𝑎𝑏 = 6𝑥 = 2(𝑥 )(3)
Therefore, we can deduce that;
𝑏2 = (3)2
Hence, we can write;
[(𝑥 )2 − 2(𝑥 )(3)] + 7
To complete the square, we can add and subtract the deduced value of 𝑏2 ;
[(𝑥 )2 − 2(𝑥 )(3) + (3)2 − (3)2 ] + 7
[{(𝑥 )2 − 2(𝑥 )(3) + (3)2} − (3)2 ] + 7
Question
The function g is defined by 𝑔(𝑥 ) = 𝑥 2 − 6𝑥 + 7 for 𝑥 > 4. By first completing the
square, find an expression for 𝑔−1 (𝑥)_and state the domain of 𝑔−1 .
Solution
We are given that;
𝑔(𝑥 ) = 𝑥 2 − 6𝑥 + 7
We use method of “completing square” to obtain the desired form. We complete the
square for the terms which involve 𝑥.
(𝑥 2 − 6𝑥 ) + 7
We have the algebraic formula;
(𝑎 − 𝑏)2 = (𝑎)2 − 2(𝑎)(𝑏) + (𝑏)2
(𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏2
For the given case we can compare the given terms with the formula as below;
𝑎2 = 𝑥 2
2𝑎𝑏 = 6𝑥 = 2(𝑥 )(3)
Therefore, we can deduce that;
𝑏2 = (3)2
Hence, we can write;
[(𝑥 )2 − 2(𝑥 )(3)] + 7
To complete the square, we can add and subtract the deduced value of 𝑏2 ;
[(𝑥 )2 − 2(𝑥 )(3) + (3)2 − (3)2 ] + 7
[{(𝑥 )2 − 2(𝑥 )(3) + (3)2} − (3)2 ] + 7
