Critical points and min and max

In summary: There is no maximum or minimum.In summary, the question is about finding the critical point and maximum or minimum of the function f(x)=(4/5), but the derivative f'(x) cannot be found because it is undefined at x=0. This means there is no critical point or maximum/minimum for the function. The correct interpretation is to consider the definition of critical point and the constraints of the function.
  • #1
christian0710
409
9

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?
 
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  • #2
christian0710 said:

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.
 
  • #3
Your original function is [itex]f(x)= x^{4/5}[/itex] (you seem to have forgotten the "x") so the derivative is [itex]f'(x)= (4/5)x^{-1/5}= \frac{4}{5x^{1/5}}[/itex].

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.
 
  • Like
Likes christian0710

Related to Critical points and min and max

What is a critical point?

A critical point is a point on a graph where the derivative is equal to zero or does not exist. This means that the slope of the graph is either flat or undefined at that point.

How do you find critical points?

To find critical points, you need to take the derivative of the function and set it equal to zero. Then, solve for the variable to find the x-values of the critical points.

What is the difference between a local and global minimum/maximum?

A local minimum/maximum is the lowest/highest point on a small section of the graph, while a global minimum/maximum is the lowest/highest point on the entire graph. A global minimum/maximum can also be referred to as an absolute minimum/maximum.

Can a function have more than one critical point?

Yes, a function can have multiple critical points. These are points where the derivative is equal to zero or does not exist, and they can occur at different locations along the graph.

How do you determine if a critical point is a minimum or maximum?

To determine if a critical point is a minimum or maximum, you can use the first or second derivative test. The first derivative test involves checking the sign of the derivative before and after the critical point. If the sign changes from positive to negative, the critical point is a maximum. If the sign changes from negative to positive, the critical point is a minimum. The second derivative test involves evaluating the second derivative at the critical point. If the second derivative is positive, the critical point is a minimum. If the second derivative is negative, the critical point is a maximum.

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