# composition functions

• Nov 24th 2006, 11:24 PM
qbkr21
composition functions
1. Find the Inverse of:

$f(x)=\frac{2x-1}{x-6}$

$f(x)=\frac{6x-1}{x-2}$

Is this Correct?

2. Find $f{\circ}g$ and $g{\circ}f$ when
$f(x)=x^3-1$ and $g(x)=\sqrt[3]{x=1}$

I got:

$(x+1)^{1/18}-1$

$\text{Thanks for the Help!!!}$
• Nov 24th 2006, 11:42 PM
CaptainBlack
Quote:

Originally Posted by qbkr21
1. Find the Inverse of:

$f(x)=\frac{2x-1}{x-6}$

$f(x)=\frac{6x-1}{x-2}$

Is this Correct?

Yes, but this would look better if you did not use $f$ for both functions, so you would have:

$f(x)=\frac{2x-1}{x-6}$

and its inverse as:

$g(x)=\frac{6x-1}{x-2}$

Also as a personel preference I would use a different variable name in this second definition so I would write:

$g(y)=\frac{6y-1}{y-2}$

is the inverse function of

$f(x)=\frac{2x-1}{x-6}$.

RonL
• Nov 24th 2006, 11:50 PM
CaptainBlack
Quote:

Originally Posted by qbkr21
2. Find $f{\circ}g$ and $g{\circ}f$ when
$f(x)=x^3-1$ and $g(x)=\sqrt[3]{x+1}$

$
(f \circ g)(x)=(g(x))^3-1=(\sqrt[3]{x+1})^3-1=(x+1)-1=x
$

and:

$
(g \circ f)(x)=\sqrt[3]{f(x)+1}=\sqrt[3]{x^3-1+1}=x
$

RonL
• Nov 25th 2006, 06:42 AM
qbkr21
Re:
Ok, but I am definatley did get those as answers I got new answers but noone of yours

1. $f{\circ}g$= $\sqrt{x+1}$

2. $g{\circ}f$= $\sqrt[6]{x^3}$

PURE CALCULATOR

$\text{In your spare time could you please tell me what I did wrong}$
• Nov 25th 2006, 07:00 AM
earboth
Quote:

Originally Posted by qbkr21
...
2. Find $f{\circ}g$ and $g{\circ}f$ when
$f(x)=x^3-1$ and $g(x)=\sqrt[3]{x=1}$

I got:

$(x+1)^{1/18}-1$

$\text{Thanks for the Help!!!}$

Hello,

first I assume that you used the shift key where you better shouldn't use it. That means your problem reads:
$f(x)=x^3-1$ and $g(x)=\sqrt[3]{x+1}$

$f{\circ}g$ means you have to calculate: $f(g(x))$. Therefore you have to plug the term of the function g in the place of the x in function f:

$f{\circ}g=f(g(x))=(\sqrt[3]{x+1})^3-1=x+1-1=x$. That's the result CaptainBlack has already told you.

Same procedure:

$g{\circ}f=g(f(x))=\sqrt[3]{x^3-1+1}=\sqrt[3]{x^3}=x$. That's the result CaptainBlack has already told you.

To be honest: I can't guess what you have done.

EB
• Nov 25th 2006, 07:05 AM
CaptainBlack
Quote:

Originally Posted by qbkr21
Ok, but I am definatley did get those as answers I got new answers but noone of yours

1. $f{\circ}g$= $\sqrt{x+1}$

2. $g{\circ}f$= $\sqrt[6]{x^3}$

PURE CALCULATOR

$\text{In your spare time could you please tell me what I did wrong}$

Earboth has explained in more detail how we think this goes, if you want
us to tell you where you went wrong, you will need to describe what you
did.

RonL
• Nov 25th 2006, 07:19 AM
qbkr21
Re:
I must have slipped up on my Calculator, thanks for the advice...
• Nov 25th 2006, 11:56 AM
topsquark
Quote:

Originally Posted by qbkr21
I must have slipped up on my Calculator, thanks for the advice...

How are you doing this on a calculator? And surely your professor wants you to work out how you got your answer?

-Dan
• Nov 25th 2006, 11:57 AM
qbkr21
I keeps getting the same answers, so here is my work:

1.
$f(g(x))$= ${(\sqrt[3]{x+1})^3}-1$
= ${(x+1)^{1/18}}-1$

$\text{and}$

2.
$g(f(x))$= $\sqrt[3]{(x^3-1)+1}$
= $(x^3)^1/6$
• Nov 25th 2006, 12:30 PM
qbkr21
Re:
Texas Instruments Voyage 200 the bling bling of All Calc's
• Nov 25th 2006, 12:38 PM
topsquark
Quote:

Originally Posted by qbkr21
I keeps getting the same answers, so here is my work:

1.
$f(g(x))$= ${(\sqrt[3]{x+1})^3}-1$
= ${(x+1)^{1/18}}-1$

$\text{and}$

The power rule is: $(x^a)^b = x^{ab}$

So
$\left ( \sqrt[3]{x+1} \right )^3 = \left ( (x + 1)^{1/3} \right )^3 = (x + 1)^{\frac{1}{3} \cdot 3} = x + 1$

Quote:

Originally Posted by qbkr21
2.
$g(f(x))$= $\sqrt[3]{(x^3-1)+1}$
= $(x^3)^{1/6}$

For the same reason as above:
$\sqrt[3]{(x^3-1)+1} = \sqrt[3]{x^3} = (x^3)^{1/3} = x^{3 \cdot \frac{1}{3}} = x$

-Dan
• Nov 26th 2006, 11:37 AM
earboth
Quote:

Originally Posted by qbkr21
I keeps getting the same answers, so here is my work:

1.
$f(g(x))$= ${(\sqrt[3]{x+1})^3}-1$
= ${(x+1)^{1/18}}-1$

...

Hello,

from your result I believe that you used the sqrt( command. Then you have calculated:

$\left( \left( (x+1)^\frac{1}{2}\right)^\frac{1}{3}\right)^\frac{ 1}{3}$ , which will indeed give your result.

Code:

((x+1)^(1/3))^3-1

You'll get the correct result.

EB
• Nov 26th 2006, 11:56 AM
qbkr21
Re:
Ok great i got $x$ for my answer, but if you were a teacher grading tests would you accept the answer given? Do you think he or she would?
• Nov 26th 2006, 12:20 PM
topsquark
Quote:

Originally Posted by qbkr21
Ok great i got $x$ for my answer, but if you were a teacher grading tests would you accept the answer given? Do you think he or she would?

If you are speaking of the answers here:

Quote:

Originally Posted by qbkr21
I keeps getting the same answers, so here is my work:

1.
$f(g(x))$= ${(\sqrt[3]{x+1})^3}-1$
= ${(x+1)^{1/18}}-1$

$\text{and}$

2.
$g(f(x))$= $\sqrt[3]{(x^3-1)+1}$
= $(x^3)^{1/6}$

No.

${(x+1)^{1/18}}-1 = \sqrt[18]{x+1} - 1$

and

$(x^3)^{1/6} = \sqrt[6]{x^3} = \sqrt{x}$

which are not the same as the correct answers.

-Dan
• Nov 26th 2006, 12:49 PM
qbkr21
Re:
$fine$