<s>[INST] Local extremes of $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$ The task is to find local extremes of $f: \mathbb R \to \mathbb R$, $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$
There is theorem that if $x_{0}$ is local extreme of $f(x)$ then $f'(x_0) = 0$
So I start from calculating derivative of $f(x)$
$$f'(x) = ((x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}})' = \frac19(x-7)^{-\frac{8}{9}}(x-2)^\frac15 + \frac15(x-2)^{-\frac45}(x-7)^\frac19 = \\ = (x-2)^\frac15(x-7)^\frac19\cdot[\frac19(x-7)^{-1}+\frac15(x-2)^{-1}] = \\ = (x-2)^\frac15(x-7)^\frac19\cdot[\frac{14x-73}{(9x-63)(5x-10)}]= (*)$$
So $(*) = 0 <=> x = 2 \lor x = 7 \lor x = \frac{73}{14}$
How can I check if any of them is local extreme?
(according to wolframalpha $\frac{73}{14}$ is local minimum and there is no local maximum)
 [/INST] for $$f'(x)$$ i have got $$f'(x)=1/45\,{\frac {14\,x-73}{ \left( x-2 \right) ^{4/5}} \left( x-7
 \right) ^{-{\frac {8}{9}}}}
$$ and for the second derivative we have
$$f''(x)=-{\frac {434\,{x}^{2}-4526\,x+15416}{2025\, \left( x-2 \right) ^{9/5}}
 \left( x-7 \right) ^{-{\frac {17}{9}}}}
$$ here you must plugg in $$x_0=\frac{73}{14}$$
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