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?
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...