# How do you find the critical points for #f(x) = 2x^(2/3) - 5x^(4/3)#?

##### 1 Answer

You calculate its first derivative and check to see where it is equal to zero or undefined.

#### Explanation:

By definition, a **critical point** of a function that can be differentiated on its domain is any point where the first derivative of said function is either **zero** or **undefined**.

This means that you need to find the first derivative of

So, differentiate the function to get

#d/dx(f(x)) = 2[d/dx(x^(2/3))] - 5d/dx(x^(4/3))#

#f^' = 2 * 2/3 * x^((2/3-1)) - 5 * 4/3 * x^((4/3-1))#

#f^' = 4/3 x^(-1/3) - 20/3 x^(1/3)#

Now, notice that one of the terms has a *negative exponent*. This means that you can write it is

#4/3x^(-13) = 4/3 * 1/x^(1/3)#

It's obvious that this expression is undefined for

Therefore,

Now check to see if you can find a point (or more) in which

#4/3 * 1/x^(1/3) = 20/3x^(1/3)#

This is equivalent to

#1/x^(1/3) = 5x^(1/3)#

You can further simplify thisequation to get

#(1/x^(1/3))^3 = (5x^(1/3))^3#

#1/x = 125x#

#1 = 125x^2#

#sqrt(1) = sqrt(125 * x^2)#

#+-1 = x * 5sqrt5 implies x = +- 1/(5sqrt(5)) = +-sqrt(5)/25#

This means that your function will have two more critical points at

So, your function

graph{2x^(2/3) - 5x^(4/3) [-4.933, 4.93, -2.466, 2.466]}