I don't understand the process for answering this question

[PinkGlasses]

In my precalculus textbook there is a question that reads, "Find f(1/2) if (f(g(x)) = (x^4 + x^2)/(1+x^2) and g(x)= 1- x^2. I couldn't figure out how to solve this problem so I looked at the answer key, which showed:

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

So, f(x) = -x + 1 and f(1/2)= -1/2+1 = 1/2.

I don't know how they got -(1-x^2) + 1 and how that translates to -x +1. I would really like to understand the process of getting the answer instead of just knowing the answer itself. I would appreciate it if anyone could help me understand.

 

[mathmari]

In my precalculus textbook there is a question that reads, "Find f(1/2) if (f(gx)) = x^4 + x^2/1+x^2 and g(x)= 1- x^2.) I couldn't figure out how to solve this problem so I looked at the answer key, which showed:

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

So, f(x) = -x + 1 and f(1/2)= -1/2+1 = 1/2.

I don't know how they got -(1-x^2) + 1 and how that translates to -x +1. I would really like to understand the process of getting the answer instead of just knowing the answer itself. I would appreciate it if anyone could help me understand.

\(\displaystyle (f o g)(x)=x^2 => (f o g)(x)=x^2-1+1 = -(1-x^2)+1\)
You know that \(\displaystyle f(g(x))= -(1-x^2)+1 => f(1- x^2)=-(1-x^2)+1 \)
Let \(\displaystyle (1-x^2)\) be equal to \(\displaystyle y\) : \(\displaystyle f(y)=-y+1 \) .

 

[Deleted member 4993]

In my precalculus textbook there is a question that reads, "Find f(1/2) if (f(g(x)) = (x^4 + x^2)/(1+x^2) and g(x)= 1- x^2.) I couldn't figure out how to solve this problem so I looked at the answer key, which showed:

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

So, f(x) = -x + 1 and f(1/2)= -1/2+1 = 1/2.

I don't know how they got -(1-x^2) + 1 and how that translates to -x +1. I would really like to understand the process of getting the answer instead of just knowing the answer itself. I would appreciate it if anyone could help me understand.

You have missed many grouping symbols in presenting your problem. Without those, it is difficult to decipher the problem

 

[PinkGlasses]

You have missed many grouping symbols in presenting your problem. Without those, it is difficult to decipher the problem

Sorry about the missing symbols, I'll add them in.

\(\displaystyle (f o g)(x)=x^2 => (f o g)(x)=x^2-1+1 = -(1-x^2)+1\)
You know that \(\displaystyle f(g(x))= -(1-x^2)+1 => f(1- x^2)=-(1-x^2)+1 \)
Let \(\displaystyle (1-x^2)\) be equal to \(\displaystyle y\) : \(\displaystyle f(y)=-y+1 \) .

Sorry, I'm still confused on where the -(1-x^2) is coming from. ):

 

[mathmari]

Sorry, I'm still confused on where the -(1-x^2) is coming from. ):

You have \(\displaystyle f(g(x))=x^2 \), right?
When you add and subtract the number 1, you have \(\displaystyle f(g(x))=x^2-1+1 \)
Now, you want that in your equation appears \(\displaystyle 1-x^2\) .
So, \(\displaystyle f(g(x))=-(-x^2)-1+1=-(-x^2+1)+1=-(1-x^2)+1 \) .

 

[PinkGlasses]

Why are you supposed to add and subtract 1 and why does the textbook say f(x) = -x +1. Wouldn't f(x) =x^2+1?

 

[mathmari]

Why are you supposed to add and subtract 1 and why does the textbook say f(x) = -x +1. Wouldn't f(x) =x^2+1?

You have \(\displaystyle f(g(x))=x^2\) , you want that you have also in the second part of the equation a relation with \(\displaystyle g(x)=1-x^2 \) .
So you have to add and subtract 1.

Then you have \(\displaystyle f(g(x))=f(1-x^2)=-(1-x^2)+1=-g(x)+1\) .
Let \(\displaystyle g(x)\) be \(\displaystyle y\),then \(\displaystyle f(y)=-y+1\) .
By setting \(\displaystyle y=x: f(x)=-x+1 \) .

 

[HallsofIvy]

In my precalculus textbook there is a question that reads, "Find f(1/2) if (f(g(x)) = (x^4 + x^2)/(1+x^2) and g(x)= 1- x^2.

You want to find f(1/2) and you are told how to find f(g(x)). Okay so what x makes g(x)= 1/2? \(\displaystyle 1- x^2= 1/2\) is the same as \(\displaystyle -x^2= -1/2\) so \(\displaystyle x= \pm\sqrt{1/2}\).

If \(\displaystyle x= \sqrt{1/2}\) then \(\displaystyle f(1/2)= f(g(\sqrt{1/2})= ((\sqrt{1/2})^4+ (\sqrt{1/2})^2)/(1+ (\sqrt{1/2})^2)= (1/4+ 1/2)/(1+ 1/2)= (3/4)/(3/2)= 1/2\).

I couldn't figure out how to solve this problem so I looked at the answer key, which showed:

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

So, f(x) = -x + 1 and f(1/2)= -1/2+1 = 1/2.

I don't know how they got -(1-x^2) + 1 and how that translates to -x +1. I would really like to understand the process of getting the answer instead of just knowing the answer itself. I would appreciate it if anyone could help me understand.

 

[JeffM]

Why are you supposed to add and subtract 1 and why does the textbook say f(x) = -x +1. Wouldn't f(x) =x^2+1?

Let's first attack it a different way. It is longer but may be clearer.

\(\displaystyle g(x) = 1 - x^2\) is given, correct?

\(\displaystyle So\ f(g(x)) = f(1 - x^2).\) Right?

\(\displaystyle Let\ u = 1 - x^2 \implies f(g(x)) = f(u).\) Simple change of variables.

\(\displaystyle f(u) = f(g(x)) = \dfrac{x^4 + x^2}{1 + x^2} = \dfrac{x^2(x^2 + 1)}{1 + x^2} = \dfrac{x^2(x^2 + 1)}{x^2 + 1} = x^2.\) Just mechanics.

But I need to express f(u) in terms of u.

\(\displaystyle u = 1 - x^2 \implies x^2 = 1 - u.\) Any problem there?

\(\displaystyle So\ f(u) = 1 - u \implies f(x) = 1 - x.\)

\(\displaystyle Thus\ f\left(\dfrac{1}{2}\right) = 1 - \dfrac{1}{2} = \dfrac{1}{2}.\)

Now what your book is doing is a shortcut.

It gets to \(\displaystyle f(g(x)) = x^2.\) But \(\displaystyle g(x) = 1 - x^2.\)

How do I translate \(\displaystyle x^2\) into something that is expressed in terms of \(\displaystyle 1 - x^2.\)

First step is that I need a negative. Simple \(\displaystyle x^2 = - (- x^2).\) You buy that?

Second step is that I need a + 1. Well if I add a 1 and simultaneously subtract a 1 that is zero, and I can add zero without changing anything.

\(\displaystyle x^2 = -(-x^2) = 0 - (-x^2) = 1 - 1 - (-x^2) = 1 - (1 - x^2).\)

This trick of putting something into an expression by adding it and subtracting it at the same time is a very common trick.

 

[PinkGlasses]

Thank you everyone who posted on this thread for taking the time to help me.

I think Jeff's long explanation was what I needed to completely understand everything. Thank you so much.