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

1 Answer
Aug 12, 2015

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 #f(x)# and check to see if you have any points in which #f^'(x) = 0# or #f^'(x)# is undefined.

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 #x=0#, since that would make the denominator of the function equal to zero.

Therefore, #x=0# is a critical point of #f(x)# because the first derivative is undefined here.

Now check to see if you can find a point (or more) in which #f^'(x) = 0#. For this to happen, you need

#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 #x = -sqrt(5)/25# and #x = sqrt(5)/25#.

So, your function #f(x) = 2x^(2/3) - 5x^(4/3)# has three critical points at #x = {-sqrt(5)/25; 0; sqrt(5)/25}#.

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