Sunday, February 26, 2012

JCCDQBHWH_FN021(b) Range of Composite Functions

 [original source unknown]

Introduction

This question is from the usual book which did not credit the source.  It comes from some unknown junior college in an unknown year.  It is a challenging question because most students are poor at finding the range of composite functions.  Furthermore, this question has a little twist: you are given the range of the composite function, but you are required to solve for something.  So you need to, in a way, work backwards and/or use inequalities (another weak point for many students).

Students are reminded of the right-to-left convention for functions in JC as well as GCE ‘A’ level examinations.  This means in ‘fg’ the  g  is done first before the  f.  There are some university professors who use a left-to-right convention, but here we do not.  So take note.

Again metacogntion and heuristics are very important and I will illustrate their use.

Stage 1:  Understanding the Problem

What is this (part of a) question about?
range of composite functions, solving for unknown

What is given in the question?
The range of  fg.

What is the question asking for?
The value of  k  that leads to the given range.

Stage 2:  Planning the strategy

What heuristics do you think can be used for this question?
· Working forwards (considering the meanings, asking “so what?” “what next?”)
· Setting up equation/inequality
· Working/thinking backwards
· Consider equivalent expressions or rephrasing the problem

Can you recall the definition of the range of a function?  The range of a composite function?

Stage 3:  Execution

Any observations that can make your job simpler?
Yes.  Observe that  g(x)  is a quadratic with positive  x2  coefficient.  So this is a parabola that looks like a happy smile.  To locate the minimum point, we can complete the square (a technique learnt in secondary school).
g(x)  =  x2 + 2x – 1  =  x2 + 2x + 12 – 12 – 1  =  (x + 1) 2 – 2
when  x = -1,  g(x) = -2.  The minimum point is (-1,-2).  So the range of  g  is all the
numbers from  -2  upwards.  i.e.  Rg = [-2, \oo ).

So what now?
With the two-stage method, suppose now we have
x  \in  Rg
That means?
x > -2
That means?
x + k + 1 > -2 + k + 1
Why do you do that?
I want to slowly manipulate the LHS to get  ln(x + k + 1)  which is  f(x).  Continuing,
ln(x + k + 1) > ln(k – 1)
f(x) > ln(k – 1)
i.e.                            Rfg = [ln(k – 1),  \oo ).
Why is there no switching in the inequality sign?
The slope of the graph of  the natural logarithm is always positive (albeit getting less steep for increasing  x).  So applying  ‘ln’  on both sides does not change the inequality.

What is the clue again?
We are told that  Rfg = [ln 3,  \oo ).  Aha!  *epiphany*  *light bulbs flashing*
ln(k – 1) must be equal to ln 3!!!  This can be solved easily!

 figure 1 – working forward and backwards

Stage 4:  Evaluation

Substituting  k = 4,  we see that    ln(x + k + 1) =  ln(x + 5)  and with  x > -2,  this will be  > ln 3  as given in the clue.

And why  x > -2?
This is because g(x) > -2, which we knew  from completing the square.  We treat the  ‘g(x)’  as the  ‘x’  when applying  f,  because this is what  fg(x)  really means.

Stage 5:  Reflection

What did we learn from solving this question?
We used metacognition to do self-monitor and self-questioning during the 5 stage problem-solving process.
We used the following heuristics.
· Working forwards (considering the meanings, asking “so what?” “what next?”)
· Setting up equation/inequality
· Working/thinking backwards
· Consider equivalent expressions or rephrasing the problem
We learned to apply the definition of the range of fg.  There are two possible methods: the one-stage method and the two-stage method.  The latter is usually better.
In the two-stage method, the range of  fg  is found by first finding the range of  g  and then applying the function  f  to it.  After the first step of finding the inequality for  g(x),  we can simply use  x  in the formula for  f.  How?  We set  x  to be in the range of  g(x) from the previous step,  then slowly manipulate the inequality until the expression for  f(x)  appears.  This will give us the range of  fg.
From the formula for the range of  fg,  we learned how to make use of the given clue to work backwards to find the unknown k.
Difficult questions can be tackled by thinking systematically and logically, and using heuristics and metacognition.  Mathematics is hard, but it is fun after you have learned it.  If you have really learned it, you become more powerful because you can use the same technique to solve all kinds of problems in future.

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

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

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.