I am supposed to find the things on which the sign of the following integral depends, $\int_{0}^{t} p'(x)((q'(x))-3) dx$,
where $p(x)=f(g(x))\,\,\,\,\,\,\,q(x)=g(f(x)) $, where $f(x)$ is a strictly increasing function and $g(x)$ is a strictly decreasing function. Now I know that $p(x)$ and $q(x)$ both are strictly decreasing, but i think the sign of the integral will depend on functions $f(x),g(x)$ and the sign of the value $t$. Am I missing something?
I couldn't come up with examples where the sign depends on functions still.
$$\frac{dp}{dx}=\frac{df}{dg}\frac{dg}{dx}<0$$ $$\frac{dq}{dx}-3=\frac{dg}{df}\frac{df}{dx}-3<-3$$ So the anti derivative $H$ is strictly increasing and: $$\int_0^t p'(x)(q'(x)-3)dx=H(t)-H(0)$$ Is positive for $t>0$, $0$ for $t=0$ and negative for $t<0$.