Question

@Zarkon checking the answer to the 2nd part of that problem..

Answer 1

So we got:
\[g'(x)=-\int_x^\infty f(t)*dt\]
\[g''(x)=\frac{d}{dx}(g'(x))=\frac{d}{dx}(\int_x^\infty f(t)*dt)=\frac{d}{dx}(\int_x^0f(t)*dt-\int_0^\infty f(t)*dt)\]

Answer 2

Sorry was off by a sign again...
\[g''(x)=\frac{d}{dx}(g'(x))=\frac{d}{dx}(-\int_x^\infty f(t)*dt)=-\frac{d}{dx}(\int_x^0f(t)*dt-\int_0^\infty f(t)*dt)\]

Answer 3

\[=-\frac{d}{dx}(\int_x^0 f(t)*dt\space )=-(-f(x))=f(x)\]

Answer 4

yes...f(x) is the answer

Answer 5

Great, thanks.. gah.. latex is a pain, i retyped g''(x) wrong again... that last minus on top is supposed to be a plus, and thanks again :)

Answer 6

A tip for you if you don't know, you can right-click the LaTeX text -> "Show Math As" -> "TeX Commands" to retrieve the original text. :)