Recent content by ninfinity

Isn't that ∞/∞? That doesn't make sense.

Homework Statement ##\int_{2}^{\infty} ue^{-u} du## The Attempt at a Solution What I did was find the family of functions described by the indefinite integral ##\int ue^{-u} du## then found the limit as b increases without bound. $$=\lim_{b\rightarrow \infty}...

I suppose this is less of a "help me with a problem" question than a question asking why something happens. All semester I have been working with vectors and vector components in my general physics class. I understand how to do it and how to solve a complex problem using this method. What I...

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...

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

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?

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

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

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...