# Critical points and min and max

1. Jul 18, 2015

### christian0710

1. The problem statement, all variables and given/known data

Hi I'm suppose to find the critical point and minimun or maximum of f(x)=(4/5)
2. Relevant equations

I have a question regarding how to interprete the results

3. The attempt at a solution

1) we start by finding f'(x)=4/(5*x^1/5)

Now my first question is this: we cannot divide by 0, so i it correct to assume that f'(x) must be undefined at x=0?

Usually we set f'(x)= 0 to find the x value, and that value is the critical point.
4/(5*x^1/5) = 0 but the only result i can get is 4=0 if i solve this equation which is surely not true.

2. Jul 18, 2015

### Ray Vickson

What is the actual definition of critical point that your book/notes uses? I have seen slightly different definitions in different sources, so the question is not empty or silly.

Also: be careful about the connection between derivatives and maxima or minima; setting a derivative to 0 is not necessarily correct when you have constraints, such as bounds on the variables.

3. Jul 18, 2015

### HallsofIvy

Staff Emeritus
Your original function is $f(x)= x^{4/5}$ (you seem to have forgotten the "x") so the derivative is $f'(x)= (4/5)x^{-1/5}= \frac{4}{5x^{1/5}}$.

A max or min for a function will be at a point where the derivative is 0 or does not exist. Yes, here the derivative does not exist at x= 0. There is no place where the derivative is equal to 0. Here it is easy to see that f(0)= 0 and f(x)> 0 for x not equal to 0.