f'(x) = 2(2/3)x-1/3 - (5/3)x2/3 = (4/3)x-1/3 - (5/3)x2/3 f"(x) = (4/3)(-1/3)x-4/3 - (5/3)(2/3)x-1/3 = (-4/9)x-4

Question

how would I set the 1st and 2nd derivatives to zero and solve for x ?

Answer 1

So, if

 

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

 

you can factor out (1/3)x^-1/3 out of each term and get:

 

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

 

then, rewrite the x with the negative exponent as the denominator of a fraction:

 

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

 

This is equal to 0 when the top of the fraction is equal to 0, i.e.:

 

4-5x = 0

4 = 5x (add 5x to both sides)

5x = 4 (swap sides for appearances)

x = 4/5 (divide both sides by 5 to get x by itself)

 

Also, f'(x) is undefined when the denominator of the rewritten version is 0, i.e:

3x^1/3 = 0

x = 0 (divide 0 by 3, and then cube both sides, you still wind up with 0 every time on the right)

 

If you're looking for critical values of the first derivative it's important to look for places where the first derivative is undefined, as well as places where it is equal to 0.

 

I'll leave the second derivative to you, but its a similar process.

 

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