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I want to graph the function $f(x) = \frac{9}{14}x^\frac{1}{3}(x^2 - 7)$.

To start, I know I need the derivative, $f'(x)$, because I want to find the critical points.

I found that $$f'(x) = \frac{3}{14}x^{-\frac{2}{3}}(x^2 - 7) + \frac{9}{7}x^{\frac{4}{3}}$$

Then, I set $f'(x) = 0$ and got $-1, 0, 1$ as my three potential critical points. I still have to verify that these are all three in fact critical points?

I also need to find my inflection points.

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  • $\begingroup$ Remember inflection points are the $x$ where $f''(x) = 0$. $\endgroup$ – Mnifldz Mar 18 '15 at 17:47
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You're given

$$f(x) \;\; = \;\; \frac{9}{14}x^\frac{1}{3}(x^2 - 7)$$

First, find a nicely factored form (if possible) for the first derivative.

$$f'(x) \;\; = \;\; \frac{9}{14}\cdot \frac{1}{3}x^{-\frac{2}{3}}\cdot(x^2 – 7) \;\; + \;\; \frac{9}{14}x^{\frac{1}{3}} \cdot 2x$$

$$f'(x) = \frac{3}{14}x^{-\frac{2}{3}}(x^2 – 7) \;\; + \;\; \frac{18}{14}x^{\frac{1}{3}} \cdot x$$

$$f'(x) \;\; = \;\; \frac{1}{14}x^{-\frac{2}{3}} \left[ 3(x^2 – 7) \;+ \; 18x^{\frac{3}{3}} \cdot x \right]$$

$$f'(x) \;\; = \;\; \frac{1}{14}x^{-\frac{2}{3}} ( 3x^2 – 21 + 18x^2)$$

$$f'(x) \;\; = \;\; \frac{1}{14}x^{-\frac{2}{3}} ( 21x^2 – 21)$$

$$f'(x) \;\; = \;\; \frac{21}{14}x^{-\frac{2}{3}} ( x^2 – 1)$$

$$f'(x) \;\; = \;\; \frac{3}{2}x^{-\frac{2}{3}} ( x^2 – 1)$$

From this we can make a sign chart for the first derivative. The "division points" will be where $f'(x)$ is zero or undefined, which gives $x = 0, \pm 1.$ For several carefully worked out examples, see this 29 October 2006 sci.math post of mine. As for the 2nd derivative, if you work things out like I did above, you should get the nice factored result $f''(x) = x^{-\frac{5}{3}}(2x^2 + 1),$ which has no zeros and is only undefined at $x=0.$

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