Reverse Composition to find composition of a derivative

[ninfinity]

Homework Statement

$$I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})$$
$$Find: g'(x-1)$$

Homework Equations

In order to find $g'(x-1)$ I know the following steps have to be taken:
$$f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)$$

The Attempt at a Solution

Composing $\sqrt{x}$ into $f(x)$ isn't hard, given that I have $f(x)$. Finding the derivative and then composing $x-1$ into that can also be done without much difficulty.
The only problem that I find myself coming across is how exactly to move from $f(x+1)$ to $f(x)$. I don't remember any sort of formal equation of something like this, so I started just playing around with it.
I defined $h(x)=x+1$ so that my original equation becomes $f(h(x))= \sqrt{x^2-2x}$.
Then I tried to logic out what had to be done to $h(x)$ to become $f(h(x))$. First I decided it needed to be squared. $$h(x)^2 = (x+1)^2 = x^2+2x+1$$It also needs to be taken to a one-half power, but doing that would negate the square, and so something else needs to be happening under the radical.
So I tried $$\sqrt{h(x)^2-1}=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x+1-1}=\sqrt{x^2+2x}$$
Unfortunately, that does not equal $f(x+1)$.
I happen to be stuck here and cannot figure out how to move forward.

 

[Saitama]

Homework Statement

$$I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})$$
$$Find: g'(x-1)$$

Homework Equations

In order to find $g'(x-1)$ I know the following steps have to be taken:
$$f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)$$

The Attempt at a Solution

Composing $\sqrt{x}$ into $f(x)$ isn't hard, given that I have $f(x)$. Finding the derivative and then composing $x-1$ into that can also be done without much difficulty.
The only problem that I find myself coming across is how exactly to move from $f(x+1)$ to $f(x)$. I don't remember any sort of formal equation of something like this, so I started just playing around with it.
I defined $h(x)=x+1$ so that my original equation becomes $f(h(x))= \sqrt{x^2-2x}$.
Then I tried to logic out what had to be done to $h(x)$ to become $f(h(x))$. First I decided it needed to be squared. $$h(x)^2 = (x+1)^2 = x^2+2x+1$$It also needs to be taken to a one-half power, but doing that would negate the square, and so something else needs to be happening under the radical.
So I tried $$\sqrt{h(x)^2-1}=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x+1-1}=\sqrt{x^2+2x}$$
Unfortunately, that does not equal $f(x+1)$.
I happen to be stuck here and cannot figure out how to move forward.

You seem to be over-thinking on the problem. The first task as you said is to find f(x).

You have f(x+1). Replace x with something which can give you f(x).

 

[ninfinity]

You have f(x+1). Replace x with something which can give you f(x).

I don't understand. Replace x with some arbitrary coefficient?

 

[Saitama]

I don't understand. Replace x with some arbitrary coefficient?

No no. Replace x with something like x+k, which turns x+1 to x.

 

[ninfinity]

No no. Replace x with something like x+k, which turns x+1 to x.

So...it would be something along the lines of f(x+k)=√x²-2x? I'm not making the connection.

 

[Saitama]

So...it would be something along the lines of f(x+k)=√x²-2x? I'm not making the connection.

Replace x with x+k, you get $f(x+k+1)=\sqrt{(x+k)^2-2(x+k)}$. What should be k if LHS is to be $f(x)$?

 

[ninfinity]

I'll admit I was still a little confused by what you meant, so I decided to try out a few numbers first. At last I tried k=-1 and noticed that inside of the function, f(x+k+1) = f(x) if k=-1, are things like this true for all composition functions and the like?

 

[Mark44]

This is obviously a calculus problem, so belongs in the Calculus & Beyond section. I am moving it there.

 

[Saitama]

I'll admit I was still a little confused by what you meant, so I decided to try out a few numbers first. At last I tried k=-1 and noticed that inside of the function, f(x+k+1) = f(x) if k=-1

Yes, k=-1 is the right choice. So what is $f(x)$ now?

are things like this true for all composition functions and the like?

Yes, they are.

 

[ninfinity]

This is obviously a calculus problem, so belongs in the Calculus & Beyond section. I am moving it there.

I apologize for that, since the question was focusing on the algebra of the problem I thought it was best suited for the precalculus section.

Yes, k=-1 is the right choice. So what is f(x) now?

$$f(x)=\sqrt{x^2-4x+3}$$

 

[Saitama]

$$f(x)=\sqrt{x^2-4x+3}$$

That's correct. You have f(x) now, do the remaining part.

 

[ninfinity]

That's correct. You have f(x) now, do the remaining part.

Then it follows that
$$f(\sqrt{x})=\sqrt{x-4\sqrt{x}+3}=g(x)$$
$$g'(x)=\frac{\sqrt{x}-2}{2\sqrt{x^2-4\sqrt{x^3}+3x}}$$
$$g'(x-1)=\frac{\sqrt{x-1}-2}{2\sqrt{(x-1)(x+2)-4\sqrt{(x-1)^3}}}$$

Those last two functions need to be simplified still further . The problem I am having, however, is that each time I go about this I seem to end up going in circles. I've been thinking about trying to factor using fractional exponents, but I'm not sure if that would get me somewhere useful. Any advice?

 

[Dick]

Then it follows that
$$f(\sqrt{x})=\sqrt{x-4\sqrt{x}+3}=g(x)$$
$$g'(x)=\frac{\sqrt{x}-2}{2\sqrt{x^2-4\sqrt{x^3}+3x}}$$
$$g'(x-1)=\frac{\sqrt{x-1}-2}{2\sqrt{(x-1)(x+2)-4\sqrt{(x-1)^3}}}$$

Those last two functions need to be simplified still further . The problem I am having, however, is that each time I go about this I seem to end up going in circles. I've been thinking about trying to factor using fractional exponents, but I'm not sure if that would get me somewhere useful. Any advice?

I don't see any terribly useful ways to simplify those. Why do you think you have to simplify?
You could factor a (x-1) out of the terms in the radical in the denominator of the second form but that really doesn't look all that much simpler.