You're almost there. What values of x make F'(x) = 0? You have three factors, so for F'(x) to be zero, at least one of the factors must be zero.TsAmE said:Homework Statement
Find the critical numbers of the function:
F(x) = x^(4/5) (x - 4)^(2)
Homework Equations
None.
The Attempt at a Solution
I differentiated and got to (1 / 5th root of x) (x - 4)(2x + 4/5(x-4))
but I don't know how I can simplify the expression to be able to solve for the critical values of x
A critical number of a function with rational exponent is a value of x where the derivative of the function is equal to zero or undefined.
To find the critical numbers of a function with rational exponent, you must first take the derivative of the function. Then, set the derivative equal to zero and solve for x. The resulting values of x are the critical numbers of the function.
Critical numbers are important in functions with rational exponent because they indicate where the function has a local maximum or minimum. This information is useful in understanding the behavior of the function and can be used to optimize the function.
Yes, a function with rational exponent can have more than one critical number. This can happen when the derivative of the function has multiple solutions for x, resulting in multiple critical numbers.
No, critical numbers and inflection points are not the same in a function with rational exponent. Inflection points occur where the concavity of the function changes, while critical numbers indicate where the derivative is zero or undefined.