Summary of what we learnt (T3 W2 - 5th - 9th JULY 2010)

Most of the class can understand what was being taught today, but some still have doubts about the equation "undefined" .
QUESTION -
Why 2/0 is undefined when 2÷0=0?

Well, firstly, note that 2/0 is the SAME as 2÷0 & hence, it would still be UNDEFINED!
Anyway, if 2 ÷ 0 is possible, we will be able to see that
letting 2 ÷ 0 = x
0 ( 2 ÷ 0) = 0( x ) .................................................................... convince yourself that 3( x ÷ 3) = x
2 = 0
This is a CONTRADICTION and hence we can see that 2 / 0 is not POSSIBLE ...

Which means that if we have 1÷x, this would be UNDEFINED if x = 0.

QUESTION -
If we have 1÷(x+1), for what value of x would render this undefinable?

For 1÷(x+1), it will be UNDEFINABLE when (x + 1) = 0 ... that means that solving it would make x = -1 ... hence, when x = -1, the function will be undefinable ... Consider the GRAPH of the function y = 1÷(x+1) (using grapher), what do you notice at x = -1?

What about 1÷(2x-3)? Hence, what would be general concept be that would make the expression 1÷(ax+b), undefinable? Does this change if the numerator was something else?

Hence, for functions of the form y = 1÷(ax+b), the DENOMINATOR can never be equal to ZERO ... and this will happen when x = -b ÷ a ... CHECK TO SEE THAT YOU KNOW WHY???

QUESTION -
In order to solve 2x - 5 = 3, we need to find the value of x, that SATISFIES (we discussed this in class today) the equation. Hence, we will have to leave x alone on the left-hand side. That means, we need to remove the constant (-5) and the coefficient (2) ...
How can this be done? I believe that Zhi Qi has a slight issue with this.

Well, when 2x - 5 = 3, we need to REMOVE the 5 on the LEFT ... hence,
2x - 5 + 5 = 3 + 5 ... NOTE that we have ADDED 5 on both sides ... leaving us with 2x = 8 ...
removing the coefficient 2 requires a DIVISION
2x ÷ 2 = 8 ÷ 2 ... leaving us with x = 4 ...

Hence, the EQUATION is SATISFIED when x = 4

CHECKING ... 2 (4 ) - 3 = 5 when is the same as the VALUE as provided in the question!

QUESTION -
When you're trying to SOLVE equations involving algebraic FRACTIONS, what is the general principal involved for questions of this form?

The general principal is to ensure that only 1 fraction is present on both sides of the equal sign!

Some are still unsure on how to use the Grapher application properly so maybe they could be provided with some instructions or Mr Ingham could briefly go through the steps with them again.

Some commented that Mr Ingham has to relax and slow down the pace of his talking as when he gets too excited or nervous he tends to talk like a bullet train :)

If you have questions that I have missed out, please ask them in class.. Thanks! :D

1 comment:

1. Before we discuss this in class, you might like to think about the following ... if we take the expression 1 / x ... what would happen to the value when

i) x = 1, 10, 100, 1000, 10000, 100000 ... hence, as x becomes very large, 1 / x will become ____________________

ii) x = -1, -10, -100, -1000, -10000, -100000 ... hence, as x becomes very small, 1/x will become _______________________

Now, if 1/x = 0 when x = 0, consider what happens when x = 0.1, 0.01, 0.001, 0.00001, ...

then consider when x = -0.1, -0.01, -0.001, -0.0001, -0.00001, ...