Let f(x) be a function defined as 

$f(x)= \begin{cases}\sin \left(x^2-3 x\right), & x \leq 0 \\ 6 x+5 x^2, & x>0\end{cases}$

Then, at $x=0, f(x)$

has a local minimum

We have,

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}} \sin \left(x^2-3 x\right)=\sin 0=0$

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0^{+}} 6 x+5 x^2=0$; and, $f(0)=0$

Clearly, $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

So, f(x) is continuous at x = 0.

Now,

In the left neighbourhood of x = 0, we have

$f'(x)=(2 x-3) \cos \left(x^2-3 x\right)<0$

$\Rightarrow f(x)$ is decreasing in the left $n b d$ of $x=0$ In the right neighbourhood of $x=0$, we have

$f'(x)=6+10 x>0$

$\Rightarrow f(x)$ is increasing in the right $n b d$ of $x=0$.

Hence, at $x=0, f(x)$ has a local minimum.