# Quod Erat Demonstrandum

## 2007/11/07

### [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 ^_^.