Help w/ composition fcns f(g(x)), g(f(x)), given f(x) = (2/x), g(x) = x^2+x-1

[Lucas3216]

Hi, good night.

I am a student who try to learn calculus alone(maybe it be a impossible task), but, let's go to the problem:

Determine the composite functions f(g(x)) and g(f(x)) and the values of x(if they exist) for which f(g(x)) = g(f(x))

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

I make this:

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

and for g(f(x)):

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

If we solve to g(f(x))

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

well, my problems just start here, because i need to make f(g(x)) = g(f(x))

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

here i made a multiplication and i got this:

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

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

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

2x^2 = -x^4 + x^3 + 7x^2 + 2x - 4 → This is correct

Ok, so now, i can't solve the problem.Anyone can help?

 

[Deleted member 4993]

Hi, good night.

I am a student who try to learn calculus alone(maybe it be a impossible task), but, let's go to the problem:

Determine the composite functions f(g(x)) and g(f(x)) and the values of x(if they exist) for which f(g(x)) = g(f(x))

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

I make this:

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

and for g(f(x)):

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

If we solve to g(f(x))

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

well, my problems just start here, because i need to make f(g(x)) = g(f(x))

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

here i made a multiplication and i got this:

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

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

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

2x^2 = -x^4 + x^3 + 7x^2 + 2x - 4 → This is correct

Ok, so now, i can't solve the problem.Anyone can help?

This equation does not have "rational roots". The only way I see to solve this is through numerical methods.

 

[Lucas3216]

This equation does not have "rational roots". The only way I see to solve this is through numerical methods.[/QUOTE

I was able to solve the equation.I made a derivative and decreased the terms, the result I get is 0,75.

 

[Deleted member 4993]

This equation does not have "rational roots". The only way I see to solve this is through numerical methods.[/QUOTE

I was able to solve the equation.I made a derivative and decreased the terms, the result I get is 0,75.

Does x = 0.75 satisfy your equation
[2x^2 = -x^4 + x^3 + 7x^2 + 2x - 4]?

 

[Steven G]

This equation does not have "rational roots". The only way I see to solve this is through numerical methods.[/QUOTE

I was able to solve the equation.I made a derivative and decreased the terms, the result I get is 0,75.

Or does 0.75 solve the derivative = 0?

Look at this example. Suppose x²-4 = 0. So x = +/- sqrt(2)

Taking the derivative of both side (of x²-4 = 0) we get 2x = 0. Then x = 0. NOT x = +/- sqrt(2). So x = 0 is WRONG.

There are many things wrong with your method. Let's just consider polynomials. Suppose p(x) has degree 3. Then p(x) can (and let's assume does) have 3 real roots.

Now p'(x) is of degree 3-1 or 2. So p'(x) can have at most 2 real roots. That is you can't get the third root from p'(x)!! In fact, for most p(x) you can't find ANY roots of p(x) by setting p'(x)=0.

It is great that you are trying to learn calculus on your own, but first you need to know your precalculus and trigonometry very well. The calculus part of calculus is actually quite easy. So why do so many students have trouble getting good grades in calculus? The answer is quite simple. Calculus requires you to know precalculus and trigonometry and many calculus students do not have that solid background.

 

[JeffM]

I am wondering whether this problem may have been transcribed incorrectly from the text. (Or perhaps there is a typo in the text.)

The type of problem is fine, but why would any teacher give one that does not have rational roots or an easily factorable quartic.

 

[Lucas3216]

Guys, sorry about that. My teacher shows me the problem this way I told you.

I will solve the problem again and pay more attention to the details.

And the book I use(Calculo:Um curso moderno e suas aplicações- LTC 7th ed (published in 2003)) gives me that problem.

Sorry for any problems i caused

 

[lookagain]

Look at this example. Suppose x²-4 = 0. So x = +/- sqrt(2)

Taking the derivative of both side (of x²-4 = 0) we get 2x = 0. Then x = 0. NOT x = +/- sqrt(2). So x = 0 is WRONG.

Your solutions to the equation of your example don't match it.
Maybe you intended the equation \(\displaystyle \ \ x^2 - 2 = 0.\)