Critical points and min and max

[christian0710]

Homework Statement

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

Homework Equations

[/B]
I have a question regarding how to interprete the results

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.

What is the correct way to interprete or argue about this result?

 

[Ray Vickson]

Homework Statement

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

Homework Equations

[/B]
I have a question regarding how to interprete the results

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.

What is the correct way to interprete or argue about this result?

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.

 

[HallsofIvy]

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.