Question 5a from the 2008 hsc is giving me a bit of difficulty: the hardest part being:
fidn the inverse for the function: f(x) = x - 1/2 x/\2
Also- for this subject, what is the raw mark needed to achieve a band E4?
i'll give it a try then.
(i)
f(x)=x-0.5x^2
=x(1-0.5x)
f(x) is a parabola with x intercepts @ x=0 and x=2
Note: f(x) is only defined for x is less than or equal to one.
then sketch the graph using the given domain. use a table of values if needed.
then sketch the graph of the inverse function. (using reflection of y=x)
(ii)
y=x-0.5x^2
x=y-0.5y^2 for (y less than or equal to one)
0.5y^2 - y + x =0
then use the quadratic formula to find y.
this gives you y = 1 plus or minus the sq root of (1-2x) but since y is less than or equal to 1 therefore y = 1 - sq root of (1-2x)
therefore inverse function = 1 - sq root of (1-2x) cause we made y the subject again.
(iii)
sub in x = 3/8 into inverse function = 1 minus sq root of (1-2x)
and the answer ends up being 0.5.
hope u can understand that.
