Find the derivative and critical numbers of a cubed root function

[frosty8688]

1. Find the intervals of increase and decrease



2. $C(x)=x^{1/3}(x+4)$



3. $C(x)=x^{4/3}+4x^{1/3}; C'(x)=\frac{4}{3}x^{1/3}+\frac{4}{3}x^{-2/3}=\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{x^{2/3}}{x^{2/3}}*\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{4x+4}{3x^{2/3}}$ I am wondering why the only critical number is -1, when 0 should also be considered.

 

[LCKurtz]

Yes, 0 should be considered. Whether it gives a relative extremum is another matter.

 

[Mark44]

1. Find the intervals of increase and decrease
2. $C(x)=x^{1/3}(x+4)$
3. $C(x)=x^{4/3}+4x^{1/3}; C'(x)=\frac{4}{3}x^{1/3}+\frac{4}{3}x^{-2/3}=\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{x^{2/3}}{x^{2/3}}*\frac{4x^{1/3}}{3}+\frac{4}{3x^{2/3}}=\frac{4x+4}{3x^{2/3}}$ I am wondering why the only critical number is -1, when 0 should also be considered.

A slightly better way to write the derivative is:
C'(x) = (4/3)x^-2/3(1 + x)

 

[frosty8688]

Unless 0 isn't considered, because it would make C(x) 0.

 

[Mark44]

Unless 0 isn't considered, because it would make C(x) 0.

Unless there are some constraints you haven't shown, it wouldn't matter that C(x) = 0.

 

[frosty8688]

Unless, because it is still increasing from -1 to 0 and from 0 to ∞.

 

[LCKurtz]

I think most calculus texts define a critical point as a point on the graph where either f'(x) = 0 or where the derivative doesn't exist. You have to include them when examining for relative extrema or changes in concavity. There's no "unless".

 

[frosty8688]

I think the answer at the back of the book was referring to the intervals it was increasing or decreasing on. That would explain it.