Showing that a differentiable function is increasing and bounded Suppose that $f:(a,b)\to\mathbb{R}$ is differentiable on $(a,b)$
and $|f'(x)|\leqslant L$
Let $g(x)=f(x)+Lx$
Show that $g:(a,b)\to\mathbb{R}$ is increaing and bounded on $(a,b)$.
Consider $x_{1},x_{2}\in(a,b)$ where $x_{1}<x_{2}$.
Now $g(x_{1})=f(x_{1})+Lx_{1}$
$g(x_{2})=f(x_{2})+Lx_{2}$
Now $g(x_{2})-g(x_{1})=f(x_{2})+Lx_{2}-f(x_{1})-Lx_{1}$
$g(x_{2})-g(x_{1})=f(x_{2})-f(x_{1})-L(x_{1}-x_{2})$
Show that $f(x_{2})-f(x_{1})\geqslant L(x_{1}-x_{2})$
$\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}\leqslant L\iff f(x_{2})-f(x_{1})\leqslant L(x_{2}-x_{1})$
Any hints on where to go next?

We have $-L\le f'\le L.$ Thus, since $g'=f'+L,$ we see $0\le g'\le 2L.$
i) $0\le g'$ implies $g$ is increasing, by the MVT.
ii) Fix $x_0\in (a,b).$  Then for any $x\in (a,b),$ $$g(x)=g(x)-g(x_0)+g(x_0) = g'(c_x)(x-x_0) + g(x_0)$$ by the MVT. This implies $|g(x)|\le L|x-x_0| + |g(x_0)| \le L(b-a)+|g(x_0)|$ for any $x\in (a,b).$ Hence $g$ is bounded on $(a,b).$
