Composite Functions-edited

[Becky4paws]

I am having problems with the second part of this problem. Instructions say "obtain the composite functions f(g(x)) and g(f(x)), and find all values (if any) of x such that f(g(x)) = g(f(x)).

Does that mean I am suppose to set the functions equal to each other and solve for x?

Problem:
f(x) = 2/x
g(x) = x^2 + x -1

My answer:

f(g(x)) = 2/x^2 + x -1
g(f(x)) = 4/x^2 + 2/x -1

Now how do I find the values of x such that f(g(x)) = g(f(x))?

 

[stapel]

Set f(g(x)) equal to g(f(x)), and solve the resulting equation, using the techniques you learned back in algebra.

Eliz.

 

[Becky4paws]

I have done that and my answer still is way off the books answer. The book says f(g(x)) = g(f(x)) if x~2.836 or x~0.705. I'm not getting that.

 

[stapel]

I have done that and my answer still is way off the books answer.

Please reply showing your steps, so that we may try to find your error.

Thank you.

Eliz.

 

[Becky4paws]

2/x^2 +x -1 = 4/x^2 +2/x -1

rewritten:
2/x^2 +x-1 = 5/x^2 +x -1

2x^2 +2x-2 = 5x^2 +5x-5
3=3x^2 +x
1=3x(x+1)

So x=1/3 or x=-1

 

[stapel]

2/x^2 +x -1 = 4/x^2 +2/x -1

Your formatting is ambiguous; I will assume that the above means the following:


. . . . .\(\displaystyle \frac{2}{x^2\,+\,x\,-\,1}\,= \,\frac{4}{x^2}\,+\,\frac{2}{x}\,-\,1\)


I would suggest converting the right-hand side to the common denominator of "x²", combining the rational terms, and then cross-multiplying the equation. Then solve the resulting quadratic.

rewritten:
2/x^2 +x-1 = 5/x^2 +x -1

I'm sorry, but I don't see how you got this...?

Please reply with clarification. Thank you.

Eliz.

 

[Becky4paws]

I guess I'm just way out of my league. Sorry for the inconvenience.

 

[stapel]

Sorry for the inconvenience.

It's not a matter of "inconvenience". It's just that it's impossible for us to help you find your error if we can't see your steps.

I apologize for my lack of clarity.

Eliz.

 

[Becky4paws]

Problem is:

2/(x^2+x-1) = (4/x^2)+(2/x) -1

I haven't done algebra in a long time. I believe I should multiply both sides by x^2, but I'm not sure if I remember the rules on doing this. I think it would work out to be 2x^2/(x^2+x-1) = 4+2x-x^2. Is this right so far?

Then do I need to multiply both sides by (x^2+x-1)? I really appreciate your help.

 

[pka]

Multiply both sides by \(\displaystyle \L
x^2 \left( {x^2 + x - 1} \right)\).

That is the L.C.D.

 

[Becky4paws]

I need to review

2x^2 = 4+2x-1

 

[pka]

\(\displaystyle \L
\begin{array}{l}
x^2 \left( {x^2 + x - 1} \right)\left[ {\frac{2}{{x^2 + x - 1}}} \right] = x^2 \left( {x^2 + x - 1} \right)\left[ {\frac{4}{{x^2 }}} \right] + x^2 \left( {x^2 + x - 1} \right)\left( {\frac{2}{x}} \right) - x^2 \left( {x^2 + x - 1} \right) \\
2x^2 = 4\left( {x^2 + x - 1} \right) + 2x\left( {x^2 + x - 1} \right) - x^2 \left( {x^2 + x - 1} \right) \\
\end{array}\)

 

[Becky4paws]

Still trying

OK - I'm unsure of how to multiply the 2/x(x^2)(x^2+x-1).

Does the denominator of x cancel out or do I get 2x^2(2x^2+2x-2)/x?

 

[pka]

You should get \(\displaystyle \L
2x\left( {x^2 + x - 1} \right)\)

 

[Becky4paws]

Here's what I got so far.

2x^2=4x^2+4x-4+2x^3+2x^2-2x-x^4-x^3-x^2

2x^2=-x^4+x^3+5x^2-2x-4.

How's that so far?