Sunday, February 26, 2012

JCCDQBHWH_FN014(c) Condition for Composite Functions

[original source unknown]



Introduction

     This (part of a) question is taken from a certain book sold in Singapore which did not credit the original sources, so I do not know which junior college or which year it is taken from.   This is a common type of question regarding condition for the existence of composite functions, with a little twist.  Many students do not feel confident solving this kind of question, which tests one’s understanding of concepts besides algebraic manipulation.  I shall show you how.

     Note that composite functions are written with a right-to-left convention i.e. the function on the right comes first.  That means in the function  gf,  f  is applied first, then g.  This seems counter-intuitive.  Think of it like this:  gf(x)  means  g(f(x))  by definition.  Start with  x.  First we apply  f.  This  gives  f(x)  i.e.  f( )  wraps around the  x.  Next, we apply  g,  so we take g( )  and wrap it around  f(x)  to get  g(f(x)).  This is like putting on a shirt/blouse and then putting on a coat.


     As usual, metacogntion and heuristics are very important for solving this question.



Stage 1:  Understanding the Problem

What is this (part of a) question about?
Condition for existence composite functions, (domain) restriction of functions

What is the question asking?
Find the least value of  k  and the value of  a  so that the function  gf  exists.

Stage 2:  Planning the approach

Have you solved a similar problem before?  How was it solved last time?
Yes, it was solved by considering the condition for existence of  gf,  interpreting their meaning, and using appropriate inequalities.  Use the “thinking forward” heuristic: keep asking “what does this mean?”  and “So what? ”.

What is different this time?
There is an additional  “x  not equal to something”  type of condition.

Do you think the same tactic can work?
Maybe.  I can try.

Stage 3:  Executing the plan

What does it mean for the composite function  gf  to exist?
It means the range of  f  (the first function)  is contained in the domain of  g.
Write:  Rf ` \subseteq ` Dg.  Here  f  means the new  f  with the restricted domain.

So what does this mean?
It means  f(x)  is a member of  the domain of  g
Write:  f(x) ` \in ` (1, ` \oo `)\{2}.   [all the numbers bigger than one, except the number 2]

So what does this mean?
It means  f(x) > 1  and  f(x) ` != ` 2.
which means  x2 + 2x > 1  and  x2 + 2x ` != ` 2.



figure 1 – working forward

We continue the line of reasoning using the technique of competing the square for the ‘>’ and ‘` != `’ inequalities.  We need to be careful when taking square roots.  Fortunately, in this situation, we know that  x + 1 > 0,  since  x > -1  (x being in the domain of f).  So we only need to consider the positive square root.  We end up with
                     x > -1 + ` \sqrt(2) `   and                  x ` != ` -1 +  ` \sqrt(3) `

Obviously,  a = -1 +  ` \sqrt(3) `  (the value that  x  is not supposed to be equal to).
The statement     x > -1 + ` \sqrt(2) `  actually implies that a whole plethora of statements
                           x > k  with  k = -1 + ` \sqrt(2) ` = 0.4142…
                           x > k  with  k = 0.5
                           x > k  with  k = 0.6
                           x > k  with  k = 1.1
                           x > k  with  k = 999
                           x > k  with  k = 9 999
                           x > k  with  k = 99 999
                           x > k  with  k = 1 000 000 000
                           … etc.
can be true.  Among these the least possible  k  is -1 + ` \sqrt(2) ` (of course!!!).



Stage 4:  Evaluation

Is it possible to check your answer?  How?

     We can store  X2 + 2X  as a function.  For example, on the TI-84, we can store the formula into function variable Y1.  We can numerically evaluate  Y1(-1 + ` \sqrt(2) `) = 1  and  Y1(-1 + ` \sqrt(3) `) = 2.  We can verify numerically that, for example,  Y1(0.5),  Y1(0.6159),  Y1(1.3546) ,  Y1(99)  … etc gives values greater than 1  i.e. values in the domain of  g.  We check through the above steps to make sure every step is correct and makes sense.


Stage 5:  Reflection

What did you learn from solving this question?
     I learned to make use of the condition  Rf ` \subseteq ` Dg  for composite function.
     I learned to work forward  by systematically refining the above statement.
     I remembered the “completing the square” technique learned from secondary school.
     I remembered being careful when dealing with square roots in inequalities.
     I learned to check my work using the calculator.

Anything else you have learned from this question?  Post your comments below.

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