Points of inflection help Find and classify all critical points and points of inflection. Use the information obtained to sketch a reasonable graph of the function.
$f(x)=x^3-9x^2+24x-14$
So far I found $$f"(x)= 6x-18$$ and factored it to $$f''(x)=6(x-3)$$ I don't know what to do after this step.
 A: The concavity part: The curve is concave up (facing up) in the region where $f''(x)\gt 0$, that is, where $x\gt 3$. The curve is concave down in the region where $f''(x)\lt 0$, that is, for $x\lt 3$. Concavity changes at $x=3$, so at $x=3$ we have a point of inflection. 
Critical points: The critical points of $f(x)$ are the points where the first derivative of $f(x)$ is $0$ (or undefined).
We have $f'(x)=3x^2-18x+24=3(x^2-6x+8)=3(x-2)(x-4)$. Thus $f'(x)=0$ at $x=2$ and at $x=4$. These are the critical points. 
Critical points are useful for identifying local maxima and minima of our function.
It is not too hard to see that $3(x-2)(x-4)\gt 0$ if $x\lt 2$, and also if $x\gt 4$. And $f'(x)\lt 0$ if $2\lt x\lt 4$.
So our function $f(x)$ is increasing in the interval $(-\infty,2)$. It is decreasing in the interval $(2,4)$, and increasing in the interval $(4,\infty)$. We reach a local maximum at $x=2$, because the function has been increasing until then, and starts to decrease. We reach a local minimum at $x=4$.  
A: solve the equation $6x-18=0$ and the calculate $f''(x)=6 \ne 0$ and we get the inflection point $x=6,y=f(6)$
A: For stationary points, find the values of $x$  that satisfy
$f'(x)=0$
For inflection points, find the values of $x$ that satisfy:
$f''(x)=0$
In the latter case, $x=3$.
