# Global maxima and global minima.

Find global maxima and global minima of $$f(x)=3(x-2)^{\frac{2}{3}}-(x-2)$$ over the interval $[0,20]$.

My input: Derivative vanishes at $x=10$ and left neighborhood gives positive derivative and right neighborhood gives negative derivative . Therefore $x=10$ is the value where function attains global maxima.(Correct me here if i write something wrong). And i am not able to figure out the global minima. Need help. I saw graph of this function at Desmos but there is a peak at $2$ i am not able to understand that too. At peak we have derivative not defined ?

You are wrong, here $x=10$ is not a global maximum point.

Notice that the function $f$ is continuous in $I=[0,20]$, but it is not differentiable at $2$. So, by Fermat's theorem, if $x_0$ is a global extremum of $f$, then one of the following is true:

• $x_0$ is a boundary point of $I$ (i.e. $x_0=0$ or $x_0=20$);
• $x_0$ is a non-differentiable point in $I$ (i.e. $x_0=2$);
• $x_0$ is stationary point in $I$, (i.e. $x_0=10$).

So you should evaluate $f$ at $0$, $2$ and $20$ and compare those values with $f(10)$ ($10$ is the stationary point that you have already found).

• Please elaborate it to me sir. Looking at graph it seems that at $x=20$ we have global minima ? Commented Jan 28, 2018 at 8:45
• See this corollary of Fermat's theorem: en.wikipedia.org/wiki/… Commented Jan 28, 2018 at 8:48
• At f(0) there is maximum value and f(2) we have minima right ? I evaluated all those value and found the at 0 we have global max. Did i do it right ? Commented Jan 28, 2018 at 9:13
• Yes, you are correct. Commented Jan 28, 2018 at 9:40

you are correct, the Maximum will be attained for $x=10$ gives $$f(10)$$ the Minimum for $x=2$ this gives $$f(2)=0$$

• If we take derivative we have (x-2) in the denominator. Don't you think at 2 it will be not defined ? Commented Jan 28, 2018 at 8:43
• the derivative yes, but $f(2)=0$ for $x=2$ we have that the derivative doesn't exists Commented Jan 28, 2018 at 8:48