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![enter image description here][1]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.

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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 $x=4$.

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  • $\begingroup$ Now do I put 3 back into the f"(x) equation to find the critical points? $\endgroup$ – Kevin Oct 6 '14 at 17:48
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    $\begingroup$ I will add some material on critical points. $\endgroup$ – André Nicolas Oct 6 '14 at 17:49
  • $\begingroup$ Note that the regions of increase, decrease give more fundamental information about the basic shape of the curve than the points of inflection. $\endgroup$ – André Nicolas Oct 6 '14 at 18:01
  • $\begingroup$ Shouldn't the critical points be (2,-4), (4,-56)? $\endgroup$ – Kevin Oct 6 '14 at 18:38
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    $\begingroup$ It depends on the conventions used in your course. Some people mention only the $x$-coordinate, and some insist on mentioning both coordinates. It sounds as if your book/instructor asks for the ordered pair. If that is the case, for where I wrote $x=2$, you should write $(2,f(2))$. By the way, I think $f(2)=6$, not $-4$. And I think your value of $f(4)$ is not right either. $\endgroup$ – André Nicolas Oct 6 '14 at 18:53
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solve the equation $6x-18=0$ and the calculate $f''(x)=6 \ne 0$ and we get the inflection point $x=6,y=f(6)$

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  • $\begingroup$ Wouldnt x=3 be the point of inflection not 6? $\endgroup$ – Kevin Oct 6 '14 at 17:51
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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, $x=3$.

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  • $\begingroup$ the condition $f''(x)=0$ gives not only inflextion points it must be $f'''(x)\ne 0$ or higher derivatives $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '14 at 17:41
  • $\begingroup$ mea culpa! thanks $\endgroup$ – Apricot Oct 6 '14 at 17:42

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