composition functions

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November 24th 2006, 10:24 PM
qbkr21

composition functions

1. Find the Inverse of:

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

My answer was:

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

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:

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

$\text{Thanks for the Help!!!}$
November 24th 2006, 10:42 PM
CaptainBlack

Quote:

Originally Posted by qbkr21

1. Find the Inverse of:

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

My answer was:

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

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
November 24th 2006, 10:50 PM
CaptainBlack

Quote:

Originally Posted by qbkr21

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

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

and:

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

RonL
November 25th 2006, 05:42 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

$\text{In your spare time could you please tell me what I did wrong}$
November 25th 2006, 06:00 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
November 25th 2006, 06:05 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
November 25th 2006, 06:19 AM
qbkr21

Re:

I must have slipped up on my Calculator, thanks for the advice...
November 25th 2006, 10:56 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
November 25th 2006, 10:57 AM
qbkr21

RE: Same Answers

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.
$g(f(x))$ = $\sqrt[3]{(x^3-1)+1}$
= $(x^3)^1/6$
November 25th 2006, 11:30 AM
qbkr21

Re:

Texas Instruments Voyage 200 the bling bling of All Calc's
November 25th 2006, 11:38 AM
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
November 26th 2006, 10:37 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.

Type on your calculator:

Code:

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

You'll get the correct result.

EB
November 26th 2006, 10:56 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?
November 26th 2006, 11:20 AM
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
November 26th 2006, 11:49 AM
qbkr21

Re:

$fine$