Let $f(x)=1+x+\frac{1}{2} x^2+\frac{1}{6}x^3$.
Let $g(x)=1+\int_0^xf(t)dt$
$h(x)=1+\int_0^xh(t)dt $.
Show that $$h(x)-f(x)=g(x)-f(x)+\int_0^x(h(t)-f(t))dt.$$
This was quite easy:
Since $h(x)-f(x)=1-f(x)+\int_0^xh(t)dt$ thus: $h(x)-f(x)=1-f(x)+g(x)-g(x)-\int_0^x(h(t)-f(t))dt$
$h(x)-f(x)=g(x)-f(x)-\int_0^xf(t)dt+\int_0^xh(t)dt=g(x)-f(x)+\int_0^x(h(t)-f(t))dt.$
Now the problem says that $h(x)-f(x)$ has a maximum at $x_0$ where $0\leq x_0\leq \frac{1}{2} $, we need to show that for $0\leq x\leq \frac{1}{2}$ $$\int_0^x(h(t)-f(t))dt\leq\frac{1}{2}(h(x_0)-f(x_0))$$
I am quite unsure how to solve this one, my guess would be to differentiate both sides. Since $\frac{d}{dx}h(x)=h(x)$, however that doesn't yield anything useful. No idea where to start. some help would be nice.
Note: I am aware that $h(x)$ is the exponential function however the problem said that we should assume non of the properties of the exponential function.
This just applies monotonicity of the integral. For any function and $a<x<b$ $$ \int_{a}^x u(t)\,dt\le \sup_{t\in[a,b]} u(t)·(x-a). $$