How do you find the intervals of increasing and decreasing using the first derivative given #y=x^(2/3)-4#?

To find the intervals of increasing and decreasing for the function ( y = x^{2/3} - 4 ), we need to analyze the sign of its first derivative ( \frac{dy}{dx} ).

1. Find the first derivative: [ \frac{dy}{dx} = \frac{2}{3}x^{-1/3} ]

2. Determine where the first derivative is positive (increasing) and where it is negative (decreasing):

   • ( \frac{dy}{dx} > 0 ) indicates increasing intervals.
   • ( \frac{dy}{dx} < 0 ) indicates decreasing intervals.

3. Solve for critical points by setting the first derivative equal to zero and finding any undefined points (where the derivative is not defined).

   Setting ( \frac{dy}{dx} = 0 ): [ \frac{2}{3}x^{-1/3} = 0 ] This implies ( x ) can't be zero, but there are no real solutions since ( x ) cannot be zero.

4. Analyze the sign of the first derivative in intervals determined by critical points and undefined points.

   Since there are no critical points or undefined points, we'll consider the sign of the first derivative over the entire domain.

   ( \frac{dy}{dx} = \frac{2}{3}x^{-1/3} ) is positive for all ( x \neq 0 ) and negative for all ( x \neq 0 ).

5. Conclusion:

   • The function ( y = x^{2/3} - 4 ) is increasing for all ( x ) except at ( x = 0 ).
   • There are no decreasing intervals for the given function.