- Joined
- Feb 16, 2005
- Messages
- 8,392
- Gender
- Male
- HSC
- 2006
Alternatively, you can justA remark on Q13c)ii).
It had a nice trap where you had to recognise that f'(x) and g'(x) are undefined at the points (1,0) and (-1,0) but these are points where the f(x) and g(x) are defined.
So technically speaking you can only say that since their derivatives are
then
Therefore you can only really claim that
This would suggest that simply substituting in say x = 1 to find c has some rigour issues as it doesn't quite fit that domain (even though it works out).
I think the intended solution was for students to recognise that since
then f(x) and g(x) share common points at x = -1 and x = 1.
From part (i), they have the same gradient behaviour in the domain -1 < x < 0 but have a common point at x = -1 on the "edge" of that domain. This means that
Similarly, they have the same gradient behaviour in the domain 0 < x < 1 but have a common point at x = 1 on the "edge" of that domain. This means that
 = g(x)\quad\text{ for }0 < x \leq 1)
- Prove f(1) = g(1) and f(-1) = g(-1) where the derivative is undefined
- Then solve for c in f(x) = g(x) + c, for one (or more) chosen x-values where the derivative is defined (e.g. x = 0.5)
