The function $f\left(x\right)=\left\{\begin{array}{l}{x}^4+x^2-x+2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for }x\le 1\\ 3x^3-x^2+x,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{for }x>1\end{array}\right.$is

$f$ being represented by polynomials is differentiable everywhere except possibly $x = 1$

$f^{\mathrm{\prime }}(1+)=\underset{h\to 0+}{lim} \frac{3(1+h{)}^3-(1+h{)}^2+(1+h)-3}{h}$

$=\underset{h\to 0}{lim} \frac{3\left[h^3+3h^2+3h\right]-\left(h^2+2h\right)+h}{h}=8$

$f^{\mathrm{\prime }}(1-)=\underset{h\to 0}{lim} \frac{(1+h{)}^4+(1+h{)}^2-(1+h)+2-3}{h}$

$=\underset{h\to 0}{lim} \frac{\left[h^4+4h^3+6h^2+4h\right]+\left(h^2+2h\right)-h}{h}=5$

$\underset{x\to 1+}{lim} f\left(x\right)=3=\underset{x\to 1}{lim} f\left(x\right)=f\left(1\right)$

So $f$ is continuous everywhere.