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Let $f(x)=3(x-2)^{\frac{2}{3}}-(x-2),~0\leq x\leq 20$ Let $x_0$ and $y_0$ be the points of the global minima and maxima, respectively, of $f(.)$ in the interval $[0,20]$. Evaluate $f(x_0)+f(y_0)$

Note that $f'(x)=2(x-2)^{-\frac{1}{3}}-1=0$ $=>x=10$ and $f''(10)=-\frac{2}{3}(10-2)^{-\frac{4}{3}}=-10.67<0$ So,at $x=10,~f(x)$ is max.

But Note that $f(10)=4$ but $f(0)=6.76$ I think I make some mistake but I can't find it out .Please solve the problem

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    Tag [tag:maxima] is intended for the questions about the software called [maxima](http://en.wikipedia.org/wiki/Maxima_%28software%29). This is stated in both tag-excerpt (which you see when you add tag to a question and when you hover above the tag with your mouse cursor) and in the [tag-wiki](http://math.stackexchange.com/tags/maxima/info). I've changed tag in your question to [tag:optimization].2012-08-08

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Hint: After finding all the local maximum and local minimum, you also need to find $f(0)$ and $f(20)$. Then compare all the values of the local maximum and local minimum with $f(0)$ and $f(20)$, then the largest one is the maximum value, and the smallest one is the minimum value.

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    Thank you all.Now I can solve this problem.2012-08-08