Suppose $f$ is increasing on $[0,\infty)$ and $\forall x: f(x)>0 \ g(x)=\frac{1}{x}\int_{0}^{x}f(u)du$ for $0
- $g(x)\le f(x)$
- $xg(x)\le f(x)$
- $xg(x)\ge f(0)$
- $yg(y)-xg(x)\le (y-x)f(y)$, $x
Suppose $f$ is increasing on $[0,\infty)$ and $\forall x: f(x)>0 \ g(x)=\frac{1}{x}\int_{0}^{x}f(u)du$ for $0
Another way to look at it:
$xg(x)$ is the total area under the curve $f(x)$ within the interval $(0,x)$. So $g(x)$ is the average height of $f(x)$ within the interval $(0,x)$. When you see $xg(x)$, think of it as an area; when you see $g(x)$, you're dealing with an average height.
So, follow your intuitive, devise various graphs or functions that violates the options.
For option 4, transform it, rearrange the terms so that it looks like something prettier and makes sense.
It looks like homework, so I won't do it completely.