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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 maxima is intended for the questions about the software called maxima. 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. I've changed tag in your question to optimization. –  Martin Sleziak Aug 8 '12 at 9:43

1 Answer 1

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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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Then Note that $f(20)=2.60$. So at $x=20$ $f(x)$ is min. But $f(2)=0$ –  Argha Aug 8 '12 at 9:51
    
@Ranabir You also have to evaluate $f(2)$ of course and compare is with the other values, as your $f$ isn't differentiable at $x= 2$. –  martini Aug 8 '12 at 10:12
    
Thank you all.Now I can solve this problem. –  Argha Aug 8 '12 at 10:19

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