Sunday, February 26, 2012

JCCDQBHWH_FN021(b) Range of Composite Functions

[original source unknown]


     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

Is the answer correct?
     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.

Any of your own reflections?  Please post in the comments below.

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.