For an equation $y = (x-3)^3 + 2x$, the first derivative is $\frac{dy}{dx} = 3(x-3)^2 + 2$. When the first derivative is 0, we have $(x-3)^2 = \frac{-2}{3}$ which is a contradiction, assuming there are only real roots. Yet for this equation, there seems to be a stationary point according to the graphic calculator that looks like this:
Does this mean there are no stationary points at all? If so, how can I sketch the curve without knowing where to "curve" in the graph?
HINT: $$f'(x)=3(x-3)^2+2\geq 2$$ for all real $x$ thus your function is strictly monotonously increasing and $$f''(x)=6(x-3)=0$$ if $x=3$ and $f''(x)=6>0$ so we have an inflection Point for $x=3,f(3)$