ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on August 20, 2009, 05:20:21 pm

Title: functions
Post by: khalil on August 20, 2009, 05:20:21 pm: If, for x>5, g(x)= In(x-5) and 2((g(x))=g(f(x)) then f(x) is equal to:
ans: x^2-10x+30, but I got (x-5)^2
Title: Re: functions
Post by: dcc on August 20, 2009, 05:25:40 pm: $2(g(x)) = \log\left( \left(x - 5\right)^2 \right)$
$g(f(x)) =\log\left( f(x) - 5\right)$

Comparing these, we find:

$f(x) - 5 = \left(x - 5\right)^2$ (check domains). This is the answer that they have given.
Title: Re: functions
Post by: TrueTears on August 20, 2009, 05:26:29 pm: $2log_e(x-5) = log_e(X)$ where $X = f(x)$

$X = (x-5)^2$

but for $g o f (x)$ remember ran f $\subset$ dom g

$\mbox{ran} f = [0, \infty)$

$\mbox{dom} g = (5, \infty)$

Thus we must restrict the domain of f so that ran f $\subset$ dom g

Thus we must move f(x) up 5 units (but remember f(x) = 5 is not included) thus the range becomes $(5, \infty)$ which is now a subset.

Thus the equation becomes $(x-5)^2 + 5$

EDIT: beaten.
Title: Re: functions
Post by: khalil on August 20, 2009, 06:32:58 pm: Quote from: TrueTears on August 20, 2009, 05:26:29 pm
$2log_e(x-5) = log_e(X)$ where $X = f(x)$

$X = (x-5)^2$

but for $g o f (x)$ remember ran f $\subset$ dom g

$\mbox{ran} f = [0, \infty)$

$\mbox{dom} g = (5, \infty)$

Thus we must restrict the domain of f so that ran f $\subset$ dom g

Thus we must move f(x) up 5 units (but remember f(x) = 5 is not included) thus the range becomes $(5, \infty)$ which is now a subset.

Thus the equation becomes $(x-5)^2 + 5$

EDIT: beaten.

How do we know what the range of f(x) is?
Title: Re: functions
Post by: TrueTears on August 20, 2009, 06:41:28 pm: That's weird why won't the LaTeX show up in my post?

Anyway there's 1 line that is missing after my first line, it doesn't show up in LaTeX

It is X = (x-5)^2

So the range is $[0, \infty)$
Title: Re: functions
Post by: khalil on August 20, 2009, 06:58:43 pm: Its as easy as 1,2,3
Title: Re: functions
Post by: khalil on August 20, 2009, 10:27:08 pm: Another

If f: [0,2pie] where f(x)= sin 2x and g:[0,2pie] g(x)=2sinx, then the value of (f+g)(3pie/2) is:

I got -2, answer is 2
Title: Re: functions
Post by: m@tty on August 20, 2009, 10:29:51 pm: I believe it should be -2
Title: Re: functions
Post by: khalil on August 20, 2009, 10:33:39 pm: Are you sure?
Its in ch 8 Q6, or if u use a diff type of essentials, its the blue part of the book the first one q)6