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ninfinity
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Homework Statement
[tex]
I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})
[/tex]
[tex]
Find: g'(x-1)
[/tex]
Homework Equations
In order to find [itex]g'(x-1)[/itex] I know the following steps have to be taken:
[tex]
f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)
[/tex]
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.