Author Topic: dcc help me thread (Read 22060 times) Tweet Share

Mao · « **Reply #30 on:** March 25, 2008, 10:01:41 pm »

this one:

Quote from: droodles on March 25, 2008, 12:01:19 am

$f(x)=\frac{x+3}{2x-1}$ with a Domain of R / (1/2)

find f o f(x)

answer is f o f(x) = x but I end up cancelling the x's all the time!

Quote from: dcc on March 25, 2008, 12:10:28 am

$f(f(x)) = \dfrac{\dfrac{x + 3}{2x - 1} + 3}{\dfrac{2x + 6}{2x - 1} - 1}$

$f(f(x)) = \dfrac{\dfrac{x + 3}{2x - 1} + \dfrac{3(2x - 1)}{2x - 1}}{\dfrac{2x + 6}{2x - 1} - \dfrac{1(2x - 1)}{2x - 1}}$

$f(f(x)) = \dfrac{x + 3 + 3(2x - 1)}{2x + 6 - 1(2x - 1)}$

$f(f(x)) = \dfrac{x + 3 + 6x - 3}{2x + 6 - 2x + 1}$

$f(f(x)) = \dfrac{7x}{7}$

$f(f(x)) = x$

Collin Li · « **Reply #31 on:** March 26, 2008, 02:24:16 am »

I wouldn't call it 'rare.' I'm sure you can cook up many examples. It's just not true that $f(f(x)) = x$ in general.

Collin Li · « **Reply #32 on:** March 26, 2008, 11:35:24 am »

Quote from: bec on March 26, 2008, 07:40:34 am

eeek sorry! i thought i remembered learning that last year but clearly not....

I think you mean $f^{-1}\left(f(x)\right) = f\left(f^{-1}(x)\right) = x$

bec · « **Reply #33 on:** March 26, 2008, 07:55:44 pm »

Quote from: coblin on March 26, 2008, 11:35:24 am

I think you mean $f^{-1}\left(f(x)\right) = f\left(f^{-1}(x)\right) = x$

yes, that would be what i was thinking of
haha i just re-read my justification of f(f(x))=0, it's quite creative

droodles · « **Reply #34 on:** March 27, 2008, 06:24:06 pm »

$g(x)=\frac {3}{(3x-4)^2} +6$
Find the smallest value of $a$ such that $f:(a,\infty) -> R, f(x)=g(x)$ is a one to one function

restricting domain yes?

dcc · « **Reply #35 on:** March 27, 2008, 06:28:33 pm »

$g(x) = \dfrac{3}{(3(x - \dfrac{4}{3}))^2} + 6 = \dfrac{1}{3(x - \dfrac{4}{3})^2} + 6$

Now this means you have an asymptote at $x = \dfrac{4}{3}$ , and since this function is symmetrical, you know that to keep this function one to one, you need to ensure that the function passes the horizontal line test.

$a = \dfrac{4}{3}$

droodles · « **Reply #36 on:** March 27, 2008, 07:14:27 pm »

find $g^-1(x)$
Answer is $g^-1(x)=\frac{4}{3}+\frac{1}{3}\sqrt\frac{3}{x-6}$
but I don't get how.

dcc · « **Reply #37 on:** March 27, 2008, 07:19:57 pm »

$f(x) = \dfrac{1}{3(x - \dfrac{4}{3})^2} + 6$

to find the inverse, you swap the x and y:

$x = \dfrac{1}{3(y - \dfrac{4}{3})^2} + 6$

$\dfrac{1}{x - 6} = 3(y - \dfrac{4}{3})^2$

$\dfrac{1}{3(x - 6)} = (y - \dfrac{4}{3})^2$

$y - \dfrac{4}{3} = \dfrac{1}{\sqrt{3(x - 6)}}$

$f^{-1}(x) = \dfrac{1}{\sqrt{3(x - 6)}} + \dfrac{4}{3}$

dcc · « **Reply #38 on:** March 27, 2008, 07:23:39 pm »

droodles · « **Reply #39 on:** March 28, 2008, 06:31:12 pm »

Just started Polynomials:

Find the values of $A$ and $B$ such that $A(x+3) + B(x+2)=4x+9$

what

Glockmeister · « **Reply #40 on:** March 28, 2008, 06:49:23 pm »

Funnily enough, I never learnt this technique in Methods. Must be assumed knowledge or something

Find the values of $A$ and $B$ such that $A(x+3) + B(x+2)=4x+9$

$A(x+3) + B(x+2)=4x+9$

Now, expand the terms in the brackets

$Ax+3A+Bx+2B=4x+9$

$(A+B)x + 3A + 2B = 4x+9$

$\implies A + B = 4 and 3A + 2B = 9$

From there, it is a simple simultaneous equation.

droodles · « **Reply #41 on:** March 28, 2008, 06:57:30 pm »

good work excal jnr thx

Mao · « **Reply #42 on:** March 28, 2008, 08:27:45 pm »

there is a simpler way to do that (second step of partical fractions):

$A(x+3)+B(x+2)=4x+9$

if we let $x=-3$

$0A+-B=-3$

$B=3$

if we let $x=-2$

$A+0B=1$

$A=1$

Glockmeister · « **Reply #43 on:** March 28, 2008, 08:36:29 pm »

Except $B = 3$ not $4$

Mao · « **Reply #44 on:** March 28, 2008, 09:23:42 pm »

Quote from: Glockmeister on March 28, 2008, 08:36:29 pm

Except $B = 3$ not $4$

oops careless arithmetics

fixed now

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