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**Open that Hampster!** $\displaystyle f(x) = x^q(x+6)$

$\displaystyle f'(x) = \frac{x+6}{5x^z} + x^!$

$\displaystyle f''(x) = \frac{5x^z + \frac{x+6}{4x^q}}{25x^k}$

Note: I couldn't get fractional exponents to show up properly, so I used this key: $\displaystyle q = \frac{1}{5} ; z = \frac{4}{5} ; k = \frac{8}{5}$

We need to find where f is concave up and down. There's two inflection points, one of which I've (correctly) found to be (0,0), which means that f is concave up at $\displaystyle (-\infty,0)$ and from $\displaystyle (s,\infty)$ and it's concave down from $\displaystyle (0,s)$.

But f''(x) never goes negative, and it's undefined at 0, so I'm not sure how to find s.