# Odd/even function and critical points

• dyn
In summary, the given function ##f(x) = 3x^{2/3}(5-x)## is an even function with a domain of all real numbers. To find its critical points, we differentiate and set the derivative equal to zero, obtaining ##10x^{-1/3}-5x^{2/3}##. This gives critical points at x = 0 and x = 2, but the expression is undefined at x = 0. By rearranging the equation, we can find a function defined for all x, giving critical points at x = 0 and x = 2. However, this raises questions about the usual rules for exponents when dealing with negative real numbers.
dyn
Homework Statement
##f(x) = 3x^{2/3}(5-x)##

Identify domain of ##f(x)## and is it odd or even or neither ?
Find the critical points of f(x)
Relevant Equations
Even function f(x) = f(-x) , odd function f(x) = -f(-x)
Critical points have the 1st derivative equal to zero
I have ##3x^{2/3}## as an even function although there is some debate as to this in another thread I started but the (5-x) factor means the function is neither odd or even. I also see the domain as all real numbers. Hopefully this is right ?
To find the critical points I differentiate f(x) to get ##10x^{-1/3}-5x^{2/3}## and set this equal to zero to get the critical points.
I can get the critical point of x=2 from this. The answer also states that x=0 is a critical point. This is the bit that confuses me.
As it stands ##10x^{-1/3}-5x^{2/3}## is not defined at x=0 but if I rearrange it as ##5x(2x^{-4/3}-x^{-1/3})## I get both critical values of x=0 and x=2.
How can the same equation be defined and not defined at the same time ? Without knowing the answer how would I know to rearrange the equation to get x=0 as an answer ?
Thanks

dyn said:
Homework Statement: ##f(x) = 3x^{2/3}(5-x)##

Identify domain of ##f(x)## and is it odd or even or neither ?
Find the critical points of f(x)
Homework Equations: Even function f(x) = f(-x) , odd function f(x) = -f(-x)
Critical points have the 1st derivative equal to zero

I have ##3x^{2/3}## as an even function although there is some debate as to this in another thread I started
I don't think there is any debate about this being an even function.
dyn said:
but the (5-x) factor means the function is neither odd or even. I also see the domain as all real numbers. Hopefully this is right ?
To find the critical points I differentiate f(x) to get ##10x^{-1/3}-5x^{2/3}## and set this equal to zero to get the critical points.
I can get the critical point of x=2 from this. The answer also states that x=0 is a critical point. This is the bit that confuses me.
As it stands ##10x^{-1/3}-5x^{2/3}## is not defined at x=0 but if I rearrange it as ##5x(2x^{-4/3}-x^{-1/3})## I get both critical values of x=0 and x=2.
The second expression isn't defined at x = 0, either, as ##x^{-1/3}## isn't defined for this value.
dyn said:
How can the same equation be defined and not defined at the same time ? Without knowing the answer how would I know to rearrange the equation to get x=0 as an answer ?
Thanks
Critical points are where the derivative is 0, or at points in the domain of the function at which the derivative is undefined.

But if I re-arrange the equation I can get a function defined for all ##x## which gives ##x=0## as an answer

dyn said:
But if I re-arrange the equation I can get a function defined for all ##x## which gives ##x=0## as an answer
No. I assume you are talking about this: ##5x(2x^{-4/3}-x^{-1/3})##, which is undefined for x = 0. It's true that 5x = 0 when x = 0, but the other factor isn't defined, which makes the whole expression (it's not an equation) undefined.

Regarding what you wrote at the end of the other thread, ##x^{1/3} = x^{2/6}##, obviously. The problem comes with the latter expression if you try to evaluate it as ##\left( x^{1/6}\right)^2## for negative x. What this shows, in my view, is that the usual rules for exponents don't carry through when you raise negative numbers to some fractional power.

dyn
Thank you Mark44 for your help. As I have previously mentioned in the other thread , it amazes me that the fractional indices for negative real numbers issue seems never mentioned in calculus of real numbers

## 1. What is an odd function?

An odd function is a function where f(-x) = -f(x) for all values of x. In other words, the function is symmetric about the origin (0,0) and has rotational symmetry of 180 degrees. This means that if you reflect the graph of an odd function across the y-axis, it will look exactly the same as the original graph.

## 2. What is an even function?

An even function is a function where f(-x) = f(x) for all values of x. This means that the function is symmetric about the y-axis and has rotational symmetry of 180 degrees. In other words, if you reflect the graph of an even function across the y-axis, it will look exactly the same as the original graph.

## 3. How do you determine if a function is odd or even?

To determine if a function is odd or even, you can use the symmetry test. Substitute -x for x in the function and see if it is equal to f(x) or -f(x). If it is equal to f(x), then the function is even. If it is equal to -f(x), then the function is odd. You can also look at the power of the highest degree term in the function. If it is an even power, the function is even. If it is an odd power, the function is odd.

## 4. What are critical points of a function?

Critical points are the points on a function where the derivative is equal to 0 or undefined. These points are important because they can tell us where the function has maximum or minimum values, or where the function changes from increasing to decreasing or vice versa. Critical points can also be used to determine the concavity of a function.

## 5. How do you find critical points of a function?

To find the critical points of a function, you need to take the derivative of the function and set it equal to 0. Solve for x to find the x-values of the critical points. Then, plug these x-values back into the original function to find the corresponding y-values. You can also use the second derivative test to determine if the critical points are maximum or minimum points.

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