Quod Erat Demonstrandum


[CE] A typical curve sketching question

Filed under: HKCEE — johnmayhk @ 9:40 下午

Just busy with the cheering team practice. For a break, write something quick…

This is a part of the question of 2000-CE-A.Math-Q.10, a typical question.Let y = f(x) = \frac{7 - 4x}{x^2 + 2}.
Show that the maximum and minimum values of f(x) are 4 and -\frac{1}{2} respectively.

When it is given to F.5C students, some of them puzzled and asked, ‘is that something wrong in the question?’ They added, ‘-\frac{1}{2} is maximum and 4 is minimum.’ How come?

In fact, students’ concept is so vague to give that comment. Saying “-\frac{1}{2} is maximum and 4 is minimum." is not clear! They should say something like:

y attains its maximum value when x = -\frac{1}{2}. Hence,
to find maximum value of y,
we put x = -\frac{1}{2} into y = \frac{7 - 4x}{x^2 + 2},
that is, the maximum value of y = \frac{7 - 4(-\frac{1}{2})}{(-\frac{1}{2})^2 + 2} = 4.

Also, they should say
y attains its minimum value when x = 4. Hence,
the minimum value of y = \frac{7 - 4(4)}{4^2 + 2} = -\frac{1}{2}.

The designer of this question played little tricks on numbers. It may be quite interesting to set up similar question by choosing ‘suitable’ numbers a and b satisfying

a < b;
f(a) = b = maximum value of y;
f(b) = a = minimum value of y; and
It is not too easy (or too difficult) to find derivatives of f(x).

Students try to find such function to help me set up A. Math Examination Paper ^_^.

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