Q:

Lipschitz condition $y = x \cdot 3^x$

I have to check if there exists a constant value $L$ for $f(x)=x \cdot 3^x$ that $f$ satisfies Lipschitz's condition on $[-5,3]$. 
So, to start with, I know that Lipschitz's condition states that $\exists L: \frac{|f(x_1) - f(x_2)|}{|x_1 - x_2|}\leq L$ on $(x_1, x_2) \in (a,b)$, so baciscally the slope of $f$ inside $(a,b)$ is finite. But how can I check that? I mean is it enough to show that $f$ is continuous inside $[-5,3]$ and takes only finite values?

A:

$f$ is continuously differentiable on the closed bounded interval $[-5,3]$. Let $x,y \in (-5,3)$ & $x<y$ (WLOG). By MVT for $f$ on $[x,y]$, there exists $c \in (x,y)$ such that $$\dfrac{f(y)-f(x)}{y-x} = f'(c).$$
Since $f$ is differentiable on $(-5,3)$, $f'(x)= 3^x(x ln(3)+1)$. $f'$ is an increasing function on $[-5,3]$ (check it!). Hence,  $f'|_{max}$ occurs at $x=3$. So, $f'(3) \geq |f'(x)|$ for all $x \in [-5,3]$. Then, $$|f(x)-f(y)| = |f'(c)||x-y| 
\leq f'(3)|x-y|$$.
From the equation for $f'$, you can calculate $f'(3)$, which is the required constant for Lipschitz condition.

